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Quantitative Techniques

This book is a part of the course by uts, Pune.This book contains the course content for Quantitative Techniques.

© uts, PuneFirst Edition 2013

The content in the book is copyright of uts. All rights reserved.No part of the content may in any form or by any electronic, mechanical, photocopying, recording, or any other means be reproduced, stored in a retrieval system or be broadcast or transmitted without the prior permission of the publisher.

uts makes reasonable endeavours to ensure content is current and accurate. uts reserves the right to alter the content whenever the need arises, and to vary it at any time without prior notice.

Published byutsBavdhan, Pune - 411021

Website : www.utsglobal.edu.inTel : +91-20-41034800, +91 9011067684

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Index

I. Content .................................................................... II

II. List of Tables..................................................... VIII

III. Application ........................................................ 105

IV. Bibliography ...................................................... 110

V. Self Assessment Answers .................................... 113

Book at a Glance

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Contents

Chapter I ....................................................................................................................................................... 1Matrices and Determinants ......................................................................................................................... 1Aim ................................................................................................................................................................ 1Objectives ..................................................................................................................................................... 1Learning Outcome: ........................................................................................................................................ 11.1 Introduction .............................................................................................................................................. 21.2 Matrix ....................................................................................................................................................... 2 1.2.1 Matrix definition ...................................................................................................................... 2 1.2.2 Matrix Notation ....................................................................................................................... 2 1.2.3 Matrix Equality ........................................................................................................................ 31.3 Types of Matrix ........................................................................................................................................ 3 1.3.1 Row Matrix .............................................................................................................................. 3 1.3.2 Column Matrix ......................................................................................................................... 3 1.3.3 Zero/Null Matrix ...................................................................................................................... 3 1.3.4 Square Matrix .......................................................................................................................... 4 1.3.5 Diagonal Matrix ....................................................................................................................... 4 1.3.6 Unit/Identity Matrix ................................................................................................................. 4 1.3.7 Transpose of a Matrix .............................................................................................................. 41.4 Operations on Matrices ............................................................................................................................ 4 1.4.1 Addition of Matrices ................................................................................................................ 4 1.4.1.1 Properties of Matrix Addition ................................................................................... 5 1.4.2 Subtraction of Matrices ............................................................................................................ 5 1.4.3 Multiplication of Matrices ....................................................................................................... 6 1.4.3.1 Multiplication of a Matrix by a Number ................................................................... 6 1.4.3.2 Multiplication of a Matrix by another Matrix ........................................................... 6 1.4.3.3 Properties of Multiplication of Matrices ................................................................... 71.5 Determinants ............................................................................................................................................ 7 1.5.1 Calculating Value of 2 x 2 Determinant ................................................................................... 7 1.5.2 Calculating Value of 3 x 3 Determinant ................................................................................... 8 1.5.2.1 Cofactors ................................................................................................................... 8 1.5.2.2 Expansion by Minors ................................................................................................ 81.6 Inverse of a Matrix ................................................................................................................................... 9 1.6.1 Finding Inverse for a 2 x 2 Matrix ........................................................................................... 9 1.6.2 Finding Inverse for a 3 x 3 Matrix ........................................................................................... 91.7 Solving Simultaneous Equation using Determinants ..............................................................................11 1.7.1 Solving Two Simultaneous Equations ....................................................................................11 1.7.2 Solving Three simultaneous Equations .................................................................................. 121.8 Properties of Determinants .................................................................................................................... 131.9 Difference between Matrices and Determinants .................................................................................... 14Summary ..................................................................................................................................................... 15References ................................................................................................................................................... 15Recommended Reading ............................................................................................................................. 15Self Assessment ........................................................................................................................................... 16

Chapter II ................................................................................................................................................... 18Mathematical Logic ................................................................................................................................... 18Aim .............................................................................................................................................................. 18Objective ...................................................................................................................................................... 18Learning outcome ........................................................................................................................................ 182.1 Introduction ............................................................................................................................................ 192.2 Definition ............................................................................................................................................... 19 2.2.1 Statement ............................................................................................................................... 19 2.2.2 Truth Value ............................................................................................................................. 19

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2.2.3 Truth Table ............................................................................................................................. 19 2.2.4 Compound Statements ........................................................................................................... 192.3 Statement ................................................................................................................................................ 202.4 Compound Statement ............................................................................................................................. 202.5 Connectives ............................................................................................................................................ 21 2.5.1 Negation ................................................................................................................................. 21 2.5.2 Conjunction ............................................................................................................................ 22 2.5.3 Disjunction ............................................................................................................................ 23 2.5.4 Conditional or Implication .................................................................................................... 23 2.5.5 Biconditional or Biimplication .............................................................................................. 24 2.5.6 Contrapositive, Converse and Inverse ................................................................................... 252.6 Tautology ,Contradiction and Contingency ........................................................................................... 252.7 Laws of Algebra ..................................................................................................................................... 26 2.7.1 Identity Law ........................................................................................................................... 26 2.7.2 Commutative Law .................................................................................................................. 26 2.7.3 Complement Law ................................................................................................................... 26 2.7.4 Double Negation .................................................................................................................... 26 2.7.5 Associative Law ..................................................................................................................... 26 2.7.6 Distributive Law .................................................................................................................... 26 2.7.7 Absorption Law ...................................................................................................................... 26 2.7.8 Demorgan’s Law .................................................................................................................... 26 2.7.9 Equivalance of Contrapositive ............................................................................................... 27 2.7.10 Others ................................................................................................................................... 27Summary ..................................................................................................................................................... 28References ................................................................................................................................................... 28Recommended Reading ............................................................................................................................. 28Self Assessment ........................................................................................................................................... 29

Chapter III .................................................................................................................................................. 31Set Theory ................................................................................................................................................... 31Aim .............................................................................................................................................................. 31Objective ...................................................................................................................................................... 31Learning outcome ........................................................................................................................................ 313.1 Definition of a Set .................................................................................................................................. 323.2 Standard Sets .......................................................................................................................................... 323.3 Representation of set .............................................................................................................................. 32 3.3.1 Tabular Form/Roaster Method ............................................................................................... 32 3.3.2 Rule Method .......................................................................................................................... 32 3.3.3 Descriptive Form ................................................................................................................... 323.4 Types of Sets .......................................................................................................................................... 33 3.4.1 Finite Set ................................................................................................................................ 33 3.4.2 Empty or Null Set .................................................................................................................. 33 3.4.3 Subset ..................................................................................................................................... 33 3.4.3.1 Proper Subset .......................................................................................................... 33 3.4.3.2 Improper Subset ...................................................................................................... 33 3.4.4 Infinite Set .............................................................................................................................. 33 3.4.5 Disjoint Sets ........................................................................................................................... 34 3.4.6 Overlapping Sets .................................................................................................................... 34 3.4.7 Universal Set .......................................................................................................................... 34 3.4.8 Equal Set ................................................................................................................................ 34 3.4.9 Complement Set ..................................................................................................................... 34 3.4.10 Equivalent Set ...................................................................................................................... 343.5 Illustration of Various Sets ..................................................................................................................... 353.6 Basic Operations on Sets ....................................................................................................................... 35 3.6.1 Intersection of Two Sets ......................................................................................................... 35

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3.6.2 Union of Two Sets ................................................................................................................. 35 3.6.3 Relative Complement or Difference of Two Sets .................................................................. 35 3.6.4 Complement of a Set .............................................................................................................. 36 3.6.5 Symmetric Difference of Two Sets ........................................................................................ 363.7 Properties of Set ..................................................................................................................................... 36 3.7.1 Commutative Law .................................................................................................................. 36 3.7.2 Associative Law ..................................................................................................................... 36 3.7.3 Distributive Law .................................................................................................................... 37 3.7.4 Identity Law ........................................................................................................................... 37 3.7.5 Complement Law ................................................................................................................... 37 3.7.6 Idempotent Law ..................................................................................................................... 37 3.7.7 Bound Law ............................................................................................................................. 37 3.7.8 Absorption Law ...................................................................................................................... 37 3.7.9 Involution Law ....................................................................................................................... 37 3.7.10 De Morgan’s Law ................................................................................................................ 37 3.7.11More Results ......................................................................................................................... 37Summary ..................................................................................................................................................... 40References ................................................................................................................................................... 40Recommended Reading ............................................................................................................................. 40Self Assessment ........................................................................................................................................... 41

Chapter IV .................................................................................................................................................. 43Progression ................................................................................................................................................. 43Aim .............................................................................................................................................................. 43Objective ...................................................................................................................................................... 43Learning outcome ........................................................................................................................................ 434.1 Introduction ............................................................................................................................................ 444.2 Arithmetic Progression ........................................................................................................................... 444.3 Formulae for Arithmetic Progression ..................................................................................................... 45 4.3.1 The general form of an AP .................................................................................................... 45 4.3.2 The nth term of an AP ........................................................................................................ 45 4.3.3 Sum of first n terms ( ) of an AP ......................................................................................... 454.4 Arithmetic Mean .................................................................................................................................... 454.5 Geometric Progression ........................................................................................................................... 454.6 Formulae for Geometric Progression ..................................................................................................... 46 4.6.1 The general form of a GP ....................................................................................................... 46 4.6.2 The nth term Tn of a GP ........................................................................................................... 46 4.6.3 The sum of first n terms Sn of a GP........................................................................................ 464.7 Geometric Mean ..................................................................................................................................... 474.8 Harmonic Progression ............................................................................................................................ 474.9 Formulae for Harmonic Progression ...................................................................................................... 48 4.9.1 The General Form of HP ........................................................................................................ 48 4.9.2 The nth term (Tn)of a HP........................................................................................................ 484.10 Harmonic mean .................................................................................................................................... 484.11 Comparison between AP and GP ......................................................................................................... 484.12 Important Rules on Arithmetic mean(AM),Geometric Mean (GM) and Harmonic Mean(HM) ......... 49Summary ..................................................................................................................................................... 51References ................................................................................................................................................... 51Recommended Reading ............................................................................................................................. 51Self Assessment ........................................................................................................................................... 52

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Chapter V .................................................................................................................................................... 54Probability .................................................................................................................................................. 54Aim .............................................................................................................................................................. 54Objective ...................................................................................................................................................... 54Learning outcome ........................................................................................................................................ 545.1 Introduction ............................................................................................................................................ 555.2 Definitions .............................................................................................................................................. 55 5.2.1 Experiment ............................................................................................................................. 55 5.2.2 Deterministic Experiment ...................................................................................................... 55 5.2.3 Random Experiment .............................................................................................................. 55 5.2.3.1 Examples of Performing a Random Experiment .................................................... 55 5.2.3.2 Details ..................................................................................................................... 55 5.2.3.3 Sample Space : ........................................................................................................ 55 5.2.4 Elementary Event ................................................................................................................... 55 5.2.5 Impossible Event .................................................................................................................... 55 5.2.6 Events ..................................................................................................................................... 56 5.2.7 Mutually Exclusive Event ...................................................................................................... 56 5.2.8 Compatibility ......................................................................................................................... 56 5.2.9 Independent Events ................................................................................................................ 56 5.2.10 Dependent Events ................................................................................................................ 565.3 Probability .............................................................................................................................................. 56 5.3.1 Probability of Occurrence of an Event .................................................................................. 56 5.3.2 Results on Probability ............................................................................................................ 56 5.3.3 Binomial Distribution ............................................................................................................ 57 5.3.4 Geometric Theorem ............................................................................................................... 575.4 Conditional Probability .......................................................................................................................... 57 5.4.1 Conditional probability of Dependent Events ....................................................................... 57 5.4.2 Conditional probability of Independent Events ..................................................................... 575.5. Multiplication Rule ............................................................................................................................... 57 5.5.1 Independent Events ................................................................................................................ 57 5.5.2 Dependent Events .................................................................................................................. 575.6 Steps to Solve Probability ...................................................................................................................... 585.7 Bayes Theorem ...................................................................................................................................... 58Summary ..................................................................................................................................................... 61References ................................................................................................................................................... 61Recommended Reading ............................................................................................................................. 61Self Assessment ........................................................................................................................................... 62

Chapter VI .................................................................................................................................................. 64Permutations and Combinations .............................................................................................................. 64Aim .............................................................................................................................................................. 64Objective ...................................................................................................................................................... 64Learning outcome ........................................................................................................................................ 646.1 Introduction ............................................................................................................................................ 656.2 Basic Calculation Used .......................................................................................................................... 65 6.2.1 Factorial Notation ................................................................................................................. 656.3 Fundamental Principles of Counting ..................................................................................................... 65 6.3.1 Principle of Addition .............................................................................................................. 65 6.3.2 Principle of Multiplication ..................................................................................................... 656.4 Permutation ............................................................................................................................................ 66 6.4.1 Basic Forms of Permutations ................................................................................................. 67 6.4.1.1 All given Objects are Distinct ................................................................................. 67 6.4.1.2 When k cannot be Selected .................................................................................... 67 6.4.1.3 When all the given n objects are not distinct .......................................................... 67 6.4.1.4 Circular Permutation .............................................................................................. 67

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6.4.1.5 Repetition is Allowed .............................................................................................. 686.5 Combination ........................................................................................................................................... 686.6 Basic Forms of Combination ................................................................................................................. 68 6.6.1All Given Objects are distinct ................................................................................................ 68 6.6.2 When K objects cannot be selected ....................................................................................... 68 6.6.3 When k Objects are always Selected .................................................................................... 69 6.6.4 Distribution of Objects into two Groups ................................................................................ 69 6.6.5 Distribution of Similar Objects .............................................................................................. 69 6.6.6 Total possible Combination of n Distinct Objects ................................................................ 69 6.6.7 When All are not Distinct Objects ......................................................................................... 70 6.6.8 When all are Distinct but of Different Kind .......................................................................... 706.7 Special Case(Permutation and Combination Simultaneously) .............................................................. 706.8 Basic Manipulation on Permutation and Combinations ........................................................................ 70Summary ..................................................................................................................................................... 73References ................................................................................................................................................... 73Recommended Reading ............................................................................................................................. 73Self Assessment ........................................................................................................................................... 74

Chapter VII ................................................................................................................................................ 76Interpolation ............................................................................................................................................... 76Aim .............................................................................................................................................................. 76Objectives .................................................................................................................................................... 76Learning outcome ........................................................................................................................................ 767.1 Introduction ............................................................................................................................................ 777.2 Definition of Interpolation ..................................................................................................................... 777.3 Application ............................................................................................................................................. 777.4 Need and Importance of Interpolation ................................................................................................... 777.5 Methods of Interpolation ........................................................................................................................ 78 7.5.1 Graphical Method .................................................................................................................. 78 7.5.2 Newton’s method of advancing differences ........................................................................... 78 7.5.3 Lagrange’s Method................................................................................................................. 78 7.5.4 Newton-Gauss Forward Method ............................................................................................ 78 7.5.5 Newton-Gauss Backward Method ......................................................................................... 79Summary ..................................................................................................................................................... 80References ................................................................................................................................................... 80Recommended Reading ............................................................................................................................. 80Self Assessment ........................................................................................................................................... 81

Chapter VIII ............................................................................................................................................... 83Consumer Arithmetic ................................................................................................................................ 83Aim .............................................................................................................................................................. 83Objectives .................................................................................................................................................... 83Learning outcome ........................................................................................................................................ 838.1 Introduction: Profit and Loss ................................................................................................................. 84 8.1.1 Formulae ................................................................................................................................ 848.2 Interest .................................................................................................................................................... 85 8.2.1 Terms Used ............................................................................................................................ 85 8.2.2 Simple Interest ....................................................................................................................... 85 8.2.2.1 Formulae ................................................................................................................. 85 8.2.3 Recurring Deposit .................................................................................................................. 86 8.2.3.1 Formulae ................................................................................................................. 86 8.2.4 Compound Interest ................................................................................................................. 86 8.2.4.1 Formulae ................................................................................................................. 86

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Summary ..................................................................................................................................................... 89References ................................................................................................................................................... 89Recommended Reading ............................................................................................................................. 89Self Assessment .......................................................................................................................................... 90

Chapter IX .................................................................................................................................................. 92Relations and Functions ............................................................................................................................ 92Aim .............................................................................................................................................................. 92Objectives .................................................................................................................................................... 92Learning outcome ........................................................................................................................................ 929.1 Relation .................................................................................................................................................. 939.2 Domain and Range of a Relation ........................................................................................................... 939.3 Functions ................................................................................................................................................ 93 9.3.1 Range, image, co-domain ...................................................................................................... 949.4 Break Even Analysis .............................................................................................................................. 94Summary ..................................................................................................................................................... 95References ................................................................................................................................................... 95Recommended Reading ............................................................................................................................. 95Self Assessment ........................................................................................................................................... 96

Chapter X .................................................................................................................................................. 98Statistics ...................................................................................................................................................... 98Aim .............................................................................................................................................................. 98Objectives .................................................................................................................................................... 98Learning outcome ........................................................................................................................................ 9810.1 Introduction .......................................................................................................................................... 9910.2 Definition of Statistics ......................................................................................................................... 9910.3 Scope and Applications of Statistics .................................................................................................... 9910.4 Characteristics of Statistics .................................................................................................................. 9910.5 Functions of Statistics ........................................................................................................................ 10010.6 Limitations of Statistics ..................................................................................................................... 10010.7 Classification ...................................................................................................................................... 10010.8 Objectives of Classification ............................................................................................................... 10010.9 Characteristics of Classification ........................................................................................................ 10010.10 Frequency Distribution .................................................................................................................... 101 10.10.1 Discrete or Ungrouped Frequency Distribution ............................................................... 101 10.10.2 Continuous or Grouped Frequency Distribution ............................................................. 101 10.10.3 Cumulative Frequency Distribution ................................................................................. 101Summary ................................................................................................................................................... 102References ................................................................................................................................................. 102Recommended Reading ........................................................................................................................... 102Self Assessment ......................................................................................................................................... 103

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List of Tables

Table 1.1 Differences between matrices and determinants .......................................................................... 14Table 2.1 Symbols of connectives ............................................................................................................... 21Table 2.2 Truth Table of Negation ............................................................................................................... 22Table 2.3 Truth table of conjunction ............................................................................................................ 22Table 2.4 Truth table for disjunction ............................................................................................................ 23Table 2.5 Truth table for implication ........................................................................................................... 24Table 2.6 Truth table of biimplication .......................................................................................................... 24Table 2.7 ∼P ∨ P is a tautology .................................................................................................................... 25Table 2.8 contradiction ................................................................................................................................. 25Table 2.9 contingency .................................................................................................................................. 25Table 4.1 Comparison between AP and GP ................................................................................................. 48

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Chapter I

Matrices and Determinants

Aim

The aim of this chapter is to:

introduce the concept of matrices•

elucidate the types of matrix•

introduce determinant of matrix•

Objectives

The objective of this chapter is to:

explicate the operations on matrices•

describe the properties of determinants•

explicate the properties of matrices•

Learning Outcome

At the end of this chapter, you will be able to:

compare different types of matrix•

identify the basic operations on matrix•

understan• d simultaneous linear equations using determinants

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1.1 IntroductionThestudyofmatricesanddeterminantsisofimmensesignificanceinthefieldofbusinessandeconomics.Thislesson introduces the matrix, the rectangular array and determinants at the heart of matrix algebra. Matrix algebra isusedquiteabitinadvancedstatistics,largelybecauseitprovidestwobenefits.

Compact notation for describing sets of data and sets of equations.•Efficientmethodsformanipulatingsetsofdataandsolvingsetsofequations.•

1.2 Matrix

1.2.1 Matrix definitionA matrix is a rectangular array of numbers arranged in rows and columns. It is a collection of real or complex numbers(usuallyreal)arrangedinafixednumberofrowsandcolumns.Itisarrangedinarectangularbrackets(either ( ) or [ ]).A set of real or complex numbers arranged in a rectangular array of ‘m’ rows and ‘n’ columns, of an order m x n (read as m by n) is called a matrix. The dimension or order of matrix is written as number of rows x number of columns.

A=

Example: A=

The topmost row is row 1.The leftmost column is column 1.

Here, the number of rows (m) is 2 and the number of columns (n) is 3.

So the matrix is of order 2 x 3 (2 by 3 matrix).

Matrices are used to solve problem in:Electronics•Statics•Robotics•Linear programming•Optimisation•Intersection of planes•Genetics•

1.2.2 Matrix Notation Statisticians use symbols to identify matrix elements and matrices.

Matrix elements: Consider the matrix below, in which matrix elements are represented entirely by symbols.•

A=

Byconventionthefirstsubscriptreferstotherownumberandthesecondsubscriptreferstothecolumnnumber.Thusthefirstelementinthefirstrowisrepresentedby ,thesecondelementinthefirstrow,by and so on, until we reach the fourth element in second row which is represented by .

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Notation: The simplest way to represent a matrix symbolically is to use bold face letters A,B ,C etc…Thus A •might refer to a 2 x 3 matrix in the below example

A =

Another approach of representing matrix A is:A= [ ] where i=1, 2 and j=1, 2, 3, 4This notation indicates A is a matrix with two rows and four columns. The actual element of the array are not displayed they are represented by the symbol .

1.2.3 Matrix EqualityTo understand matrix algebra, we need to understand matrix equality. Two matrices are equal if all three of the following conditions are met:

each matrix has same number of rows•each matrix has same number of columns•corresponding elements within each matrix are equal•

Consider the three matrices given below A= B= C=

If A=B, then x=22 and y=33, as corresponding elements of equal matrices are equal. And it is clear that C is not equal to A or B, because C has more columns than A or B.

1.3 Types of MatrixThere are six types of matrices. They are as follows.

1.3.1 Row MatrixA matrix having a single row is called row matrix.

Example

A=

1.3.2 Column MatrixA matrix having a single column is called column matrix.

Example:

A=

1.3.3 Zero/Null MatrixA matrix having each and every element as zero is called a null or zero matrix.

Example

A=

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1.3.4 Square MatrixA matrix having equal number of rows and columns is called a square matrix.

Example

A=

1.3.5 Diagonal MatrixA square matrix having all elements zero except principal diagonal elements is called diagonal matrix. Principal diagonal elements can be any non-zero elements.

Matrix elements like , , etc… are called principal diagonal elements.

Example

A=

1.3.6 Unit/Identity MatrixA square matrix which is a diagonal matrix having all principal diagonal elements as one (unit) is called identity matrix.

Example

A=

1.3.7 Transpose of a MatrixThetransposeofonematrixisobtainedbyusingtherowofthefirstmatrixasthecolumnofthesecondmatrix.

Example: if A= , then the transpose of A is represented by A’

A’=

1.4 Operations on MatricesLike ordinary algebra, matrix algebra has operations addition, subtraction and multiplication.

1.4.1 Addition of MatricesTwo matrices can be added only if they have same dimensions; that is, they must have same number of rows and columns.

Addition can be accomplished by adding corresponding elements.

For example, consider matrix A and matrix B

A= B=

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Both matrices have the same number of rows and columns (2 rows and 3 columns), so they can be added. Thus,

A+B =

A+B=

Andfinally,notethattheorderinwhichthematricesareaddeddoesnotaffectthefinalresult.A+B=B+A.

1.4.1.1 Properties of Matrix AdditionThe properties of addition of matrices are as follows:

Commutative property is true ;that is A+B=B+A• Associative property is true•A+ (B+C) = (A+B) +CDistributive property is true•K (A+B) =ka+kb(A+B) k=Ak+BkExistence of additive identity element, if a matrix is added with null matrix of the same dimension then, it results •in the same matrix, so the additive identity of a matrix is null matrixA+0=0+A=AExistence of additive inverse, if a matrix is added by inverse of A matrix, then the result is a null matrix, so the •additive inverse of a matrix is the inverse of the matrix itself matrixA+ (-A) = (-A) +A= 0

1.4.2 Subtraction of MatricesLike addition of matrices, subtraction of matrices also follows the same conditions and procedures for subtracting two matrices. Two matrices can be subtracted only if they have same dimensions; that is, they must have same number of rows and columns.

Subtraction can be accomplished by adding corresponding elements.

For example, consider matrix A and matrix B

A= B=

Both matrices have the same number of rows and columns (3 rows and 2 columns), so they can be subtracted. Thus,

A-B=

A-B=

Andfinally,notethattheorderinwhichthematricesaresubtractedaffectsthefinalresult.A-B≠ B-A.

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1.4.3 Multiplication of MatricesIn matrix multiplication there are two types of matrix multiplication. They are:

Multiplication of a matrix by a number•Multiplication of a matrix by another matrix•

1.4.3.1 Multiplication of a Matrix by a NumberWhen a matrix is multiplied by a number, every element in the matrix should be multiplied by that same number. This operation produces a new matrix, which is called scalar multiple. This multiplication process is called as scalar multiplication.

For example, if x is 5 and matrix A is as follows,

A=

Then,

xA = 5A = 5 = = = B (say)

In the example above, every element of A is multiplied by 5 to produce the scalar multiple, B.

1.4.3.2 Multiplication of a Matrix by another MatrixThematrixproductABisdefinedonlywhenthenumberofcolumnsinAisequaltothenumberofrowsinB.Similarly,thematrixproductBAisdefinedonlywhenthenumberofcolumnsinBisequaltonumberofrowsinA.

Suppose that A is an i x j matrix and B is a j x k matrix. Then, the matrix product AB results in a matrix C which has i rows and k columns; and each element in C can be computed according to the following formula.

=

Where,

= the element in row i and column k in matrix C

= the element in row i and column j in matrix A

= the element in row j and column k in matrix B

= summation sign, which indicates that the should be summed over j

Suppose we want to compute AB, given the matrices below.

A= B=

Let AB = C.Because A has 2 rows, we know that C will also have 2 rows; and because B has 2 columns, we know that C will have 2 columns. To compute the value of every element in 2 x 2 matrix C, we use the formula

= , such that

• = = 0 * 6 + 1 * 8 + 2 * 1 = 0+8+2 = 10

• = = 0 * 7 + 1 * 9 + 2 * 2 = 0+9+4 = 13

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• = = 3 * 6 + 4 * 8 + 5 * 1 = 18+32+5 = 55

• = = 3 * 7 + 4 * 9 + 5 * 2 = 21+36+10 = 67

Therefore AB= C =

1.4.3.3 Properties of Multiplication of MatricesThe properties of multiplication of matrices are as follows:

Commutative property is not true ;that is ,even when matrix multiplication is possible in both direction the •results may be different ,that is AB is not always equal to BAAssociative property is true•A (BC) = (AB) CDistributive property is true•K (AB) = (ka) (kb)(AB) k= (Ak) (Bk)Existence of multiplicative identity element, if a matrix is multiplied with identity matrix of the same dimension •then, it results in the same matrix, so the multiplicative identity of a matrix is identity matrixAI=IA=AExistence of multiplicative inverse, if a matrix is multiplied by the inverse of it, then the result is a identity •matrix, so the multiplicative inverse of a matrix is its inverse matrix (inverse of a matrix is discussed in 1.6)A * = * A = I

1.5 DeterminantsA determinant is a square array of numbers (written within a pair of vertical lines) which represents a certain sum of products.

example of a 2 x 2 determinant:

A=

1.5.1 Calculating Value of 2 x 2 DeterminantIngeneralweneedtofindthevalueof2x2determinantswithelementsa,b,canddasfollows:

= ad-cb

Herethediagonalsaremultiplied(topleft*bottomrightfirst)andthensubtracted.

ExampleFind the value of the determinant

= 4 * 3 - 2 * 1 = 12 – 2 =10 (answer)

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1.5.2 Calculating Value of 3 x 3 DeterminantA general representation of a 3 x 3 matrix is as follows

A=

Themethodusedforfindingthedeterminantsof3x3istheexpansionbyminors.

1.5.2.1 CofactorsThe 2 x 2 determinant is called the cofactor of for 3 x 3 matrix

The cofactors are formed from the elements that are not in the same row and not in the same column as .

Thus the elements in grey are not in the row and column of , so is the cofactor of .

Similarly for , the cofactor is

And for , the cofactor is

1.5.2.2 Expansion by MinorsThe3x3determinantvaluesareevaluatedbyexpansionbyminors.Thisinvolvesmultiplyingthefirstcolumnofthedeterminantwiththecofactorofthoseelements.Themiddleproductissubtractedandthefinalproductisadded.

= - +

example: evaluate = -2 – (5) + 4 = -2[(-1) (2)-(-8) (4)] – 5 [(2) (3) – (-8) (-1)] + 4 [(3) (4)-(-1) (-1)] = -2(30)-5(-2) +4(11) =-60+10+44 = -6

Hereweareusingfirstcolumntoexpandit,evenifweusefirstrowtoexpand,itgivesthesameresult.

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1.6 Inverse of a MatrixSuppose A is an n x n matrix, denoted by ,thatsatisfiesthefollowingcondition

A = A = I

Where I is the identity matrixTo check whether inverse of the matrix exists:Find the determinant of the square matrix, if the determinant value is zero then the inverse of that matrix does not exist and that matrix is known as Singular matrix.

If the determinant value is not zero, then there exists an inverse for that matrix.

The matrix for which there is an inverse is called non-singular matrix or invertible matrix.

1.6.1 Finding Inverse for a 2 x 2 MatrixSuppose A is a non-singular 2 x 2 matrix .Then, the inverse of A can be computed as given below,

A= then = |A| is the determinant value of the matrix

How tofind the determinant value of 2 x 2matrix and 3 x 3matrix are discussed above in 1.5.1 and 1.5.2respectively.

Example:

Find the inverse of the 2 x 2 matrix B =

|B| = 4 (Refer 1.5.1)

= = is the inverse of the matrix

1.6.2 Finding Inverse for a 3 x 3 MatrixStepsforfindingtheinverseof3x3matrix:

Find the determinant of a 3 x 3 matrix, det(A)•Find the transpose of the matrix•Find the determinant of the cofactors of each element in the transpose matrix.•Represent these values as a matrix of the cofactors•Find the adjoint of that resultant matrix adj(A)•Substitute the required values in •

= adj (A)Verify by multiplying A and • ,the result should be an identity matrix of same dimension.

ExampleFind the inverse of A= Step 1:Find determinant of the 3 x 3 matrix (refer 1.5.2)det (A) = 1(0-24)-2(0-20)+3(0-5)det (A)=1

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Step 2:Find the transpose of the matrix

=

Step 3:Find the determinant of the cofactor of each element in the transpose matrix.

= = -24

= = -18

= = 5

= = -20

= = -15

= = 4

= = -5

= = -4

= = 1

Step 4:Represents these values as a matrix of the cofactors

Step 5:Find the adjoint of the matrix

adj (A)=

=

Step 6:Substitute the values in

= adj (A)

=

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Therefore =

Step 7:VerificationA = I

=

If a matrix is multiplied with its inverse, then the result should be the identity matrix of same dimension

1.7 Solving Simultaneous Equation using Determinants

1.7.1 Solving Two Simultaneous EquationsSystem of equation can be solved using determinants with cramer’s rule

The solution of (x, y) of the system

x + y = ----------- (1)

x + y = ----------- (2)

can be found using determinants

Solution:Here, x and y are the variables, & arethecoefficientsofthevariablexinequations1and2respectivelyand & arethecoefficientsofthevariableyinequations1and2respectively, and are the constants of equation 1 and 2 respectively.

Step 1:Solvethedeterminantofcoefficientsofvariablesanditisrepresentedby∆

∆ =

Step 2:Solvethedeterminantreplacingconstantsinsteadofcoefficientofvariablexanditisrepresentedby

=

Step 3:Solvethedeterminantreplacingconstantsinsteadofcoefficientofvariableyanditisrepresentedby

=

Step 4:Obtain solution as x= and y =

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ExampleSolve the system using Cramer’s rule.x-3y=62x+3y=3

Solution:Here = 1; = -3; = 2; = 3; = 6; = 3;

So, x= = = = 3

y= = = = -1

So, the solution is (3,-1)

1.7.2 Solving Three simultaneous EquationsSystem of equation can be solved using determinants with cramer’s rule

The solution of (x, y, z) of the system x + y+ z = ----------- (1)

x + y+ z = ----------- (2)

x + y+ z = ----------- (3)

can be found using determinants

SolutionHere, x, y and z are the variables. , & arethecoefficientsofthevariablexinequations1,2and3respectively. , & arethecoefficientsofthevariableyinequations1,2and3respectively. , and are the coefficientsofthevariablezinequations1,2and3respectively. , and are the constants of equation 1,2 and 3 respectively.

Step 1:Solvethedeterminantofcoefficientsofvariablesanditisrepresentedby∆

∆ =

Step 2:Solvethedeterminantreplacingconstantsinsteadofcoefficientofvariablexanditisrepresentedby

=

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Step 3:Solvethedeterminantreplacingconstantsinsteadofcoefficientofvariableyanditisrepresentedby

=

Step 4:Solvethedeterminantreplacingconstantsinsteadofcoefficientofvariablezanditisrepresentedby

=

Step 5:Obtain solution as x= ; y = ; z =

ExampleSolve the system using Cramer’s rule.2x+3y+z=2-x+2y+3z=1-3x-3y+z=0

Solution:Here = 2; = 3; = 1; = -1; = 2; = 3; = -3; =-3; = 1 and =2; = 1; =0

∆ = = 2(11) +1(6)-3(7) = 7

So, x= = = 28 / 7=4

y= = = - 21/ 7 = -3

z= = = 21/ 7 = 3

So, the solution is (4,-3, 3)

1.8 Properties of DeterminantsThe properties of determinants are as follows:

The value of determinant remains unchanged if its rows and columns are interchanged•If any two rows/columns change by minus sign only ,then also the value of determinant remains unchanged•If any two rows/columns of a determinant are identical, then the value of determinant is zero•If each element of a row/column of a determinant is multiplied by a same constant and then added to corresponding •elements of some other row/column, then the value of determinant remains unchanged.If each element of a row/column of a determinant is zero, then the value of the determinant is zero.•

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1.9 Difference between Matrices and DeterminantsFollowing is the difference between matrices and determinants

Features Matrices Determinants

Definition A matrix is an array of numbers arranged in rectangular brackets.

A determinant is a square array of numbers (written within a pair of vertical lines) which represents a certain sum of products.

Representation It is written inside brackets either ( ) or [ ].

It is written within two vertical Lines | | .

Value/Result It results in an array of number inside brackets.

It results in a single number.

Influence Scalar multiplication affects all the elements in a matrix.

Scalar multiplication only affects single row /single column.

Value Matrices contain many elements.

Determinant has a single number as a end result.

Nature Matrices may positive or negative.

Determinant value is always positive. Though it results in a negative number we consider it as positive because determinant is like distance(it cannot e negative whether it is forward or backward)

Table 1.1 Differences between matrices and determinants

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SummaryA • matrix is a rectangular array of numbers arranged in rows and columns.A set of real or complex numbers arranged in a rectangular array of ‘m’ rows and ‘n’ columns, of an order m x •n (read as m by n) is called a matrix.Two matrices can be added only if they have same dimensions.•Commutative property, Associative property, Distributive property is true for matrix addition.•ThematrixproductABisdefinedonlywhenthenumberofcolumnsinAisequaltothenumberofrowsinB.•Commutative property is not true for matrix multiplication.•Associative property ,Distributive property are true for matrix multiplication•There exists additive identity, multiplicative identity, additive inverse and multiplicative inverse for a matrix.•A determinant is a square array of numbers.•The 3 x 3 determinant values are evaluated by expansion by minors.•System of equation can be solved using determinants with Cramer’s rule.•

ReferencesDr. Kala, V. N. & Rana, R., 2009. • Matrices, 1st ed., Laxmi Publication ltd.Jain, T. R. & Aggarwal, S. C., 2010. • Business Mathematics and Statistics, V.K Enterprises.Matrices and determinants• , [pdf] Available at: < http://www.kkuniyuk.com/M1410801Part1.pdf > [Accessed 31 August 2012].Gunawarden, J., • Matrix algebras for beginning, [Online] Available at: < http://vcp.med.harvard.edu/papers/matrices-1.pdf > [Accessed 31 August 2012].2011, • Matrices, [Video Online] Available at: < http://www.youtube.com/watch?v=9tFhs-D47Ik > [Accessed 31 August 2012].Hurst, W., • Matrices & determinants, [Video Online] Available at: < http://www.youtube.com/watch?v=havr-W8IwKs > [Accessed 31 August 2012].

Recommended ReadingMcMahon, D., 2005. • LinearAlgebraDemystified, McGraw-hill publication.Anton, H., 2010. • Elementary Linear Algebra, 10th ed., FM Publications.Greub, W., 1975. • Linear Algebra graduate texts in mathematics, Springer.

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Self AssessmentIf matrix A= 1.

and aij is the element of matrix A in ith row and jth column, then what is the value of a21?

3a. 4b. 2c. 5d.

It is given that P= 2. and Q = .What is the value of x+y if P=Q?3a. 5b. 6c. 8d.

A _3. ______ is a rectangular array of numbers arranged in rows and columns.Determinanta. Matrixb. Arrayc. Transposed.

Two matrices can be added only if they have __________.4. same dimensionsa. different dimensionsb. plus signc. minus signd.

When a matrix is multiplied by a number, then the process is called as _________.5. matrix multiplicationa. scalar multiplicationb. square multiplicationc. rectangular multiplicationd.

What type of matrix is A=6. ?squarea. diagonalb. nullc. identityd.

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What is the addition matrix of the following two matrices?7.

A = and B=

a.

b. c.

d.

If A = 8. , what is the value of 5A?

a.

b.

c.

d.

What is the value of determinant 9. ?6a. 8b. 7c. 10d.

A _________ is a square array of numbers. 10. matrixa. determinantb. arrayc. transposed.

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Chapter II

Mathematical Logic

Aim

The aim of this chapter is to:

introduce mathematical logic•

describe operations on logic•

highlight tautology and contradiction•

Objective

The objectives of this chapter are to:

explicate logical connectives•

elucidate laws of algebra of propositions•

describe compound statement•

Learning outcome

At the end of this chapter, you will be able to:

identify the use of mathematical logic•

understand the complex procedures into simpler form•

understand statement a• nd the truth table

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2.1 IntroductionMathematical Logic is a tool for providing a precise meaning to mathematical statements.

It includes:A formal language for expressing them•A concise notation to represent them•A methodology for objectively reasoning about their truth or falsity•

2.2 DefinitionThe part of mathematics concerned with the study of formal languages, formal reasoning, the nature of mathematical proof, provability of mathematical statements, computability, and other aspects of the foundations of mathematics.

2.2.1 StatementA statement is a declarative sentence which is either true or false but not both.

2.2.2 Truth ValueThe truth value of a proposition is true (T) if it is of true proposition and false (F) if it is false proposition.

ExampleP: The year 1973 was a leap year is a proposition readily decidable as false.

Note that the use of label ‘P…’ so that the overall statement is read p is the statement:”The year 1973 was a leap year”.

So we use P, Q, R, S, T to represent statements and these letters are called as statement variables, that is, variable replaced by statements.

ExampleDetermine whether the following sentences are statements are not.If it is a statement, determine its truth value.

The sun rises in west. False128= 26 False

Is 2 an integer? Not a statement as it is interrogative Take the book not a statement

2.2.3 Truth TableA table that gives the truth value of the compound statement in terms of its component part is called a truth table.

2.2.4 Compound StatementsA compound statement is a combination of two or more statements.

ExampleToday is Friday and it is a holiday

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2.3 StatementA statement is an assertion that can be determined to be True or False. A statement/simple statement or proposition is a declarative sentence that is either True or False but not both. A simple statement is the basic building block of the logic.

Those declarative statements will be admitted in the object language which have one and only one of two •possible values called “Truth Value”The two truth values are true and false, which are denoted by T and F respectively•Occasionally they are represented using symbols 1 and 0•We do not use other kind of statements in object language such as exclamatory and interrogative•Declarative sentences in object language are of two types•Thefirsttypeincludesthosesentenceswhichareconsideredtobeprimitiveintheobjectlanguage•This will be denoted by distinct symbols selected from uppercase letters A, B... P, Q...•Second type are obtained from the primitive ones by using certain symbols called connectives and certain •punctuation marks such as parentheses to join primitive sentences

In any case, all declarative sentences to which it is possible to assign one and only of the two possible truth values are called statements.

The following are the statements which do not contain any connectives, these kinds of statements are called as atomic or primary primitive statement.

Canada is a country1. Moscow is the capital of spain2. This statement is false3. 1+101=1104. Close the door5. Toronto is an old city6. Man will reach mars by 20807.

The statements are discussed below

The statements 1 and 2 have truth values true or false•Sentence3isnotastatementaccordingtothedefinition,becausewecannotassigntoitadefinitetruthvalue•If we assign a value true then the statement 3 is false, if assigned false then the statement 3 is true•Sentence 4 is a statement; if the numbers are considered as decimal system then the statement is false. If it is •considered as binary number system, then the statement is true. So the statement 4 is true.Statement 5 is not a statement as it is interrogative•Statement 6 is considered true in some part of the world and false in certain other parts of the world•The statement 7 could not be determined ,it will be determined only in the year or earlier when man reaches •mars before that date

2.4 Compound StatementA statement represented by a single statement variable (without any connective) is called a simple (or primitive) statement.

A statement represented by some combination of statement variables and connectives is called a compound statement.

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ExampleA dog or a car is an animalA dog is not an animal5<3Iftheearthisflat,then3+4=7

2.5 ConnectivesIn case of simple statements, their truth values are fairly obvious. It is possible to construct rather complicated statements from simpler statements by using certain connecting words or expressions known as “sentential connectives”. The statements which we initially consider are simple statements, called atomic or primary statements. New statement can be formed from atomic statements through the use of sentential connectives. The resulting statement is called molecular or compound statements. Thus the atomic statements are those which do not have any connectives. Capital letters are used to denote statements.

The capital letters with or without subscripts, will also be used to denote arbitrary statements. In the sense, a statement “P” either denotes a particular statement or serves as a place holder for any statement .This dual use of the symbol todenoteeitheradefinitestatement,calledaconstant,oranarbitrarystatementcalledavariable.The truth value of “P” is the truth table of actual statement which it represents.It should be emphasises that when “P” is used as a statement variable, it has no truth value and such does not represent a statement in symbolic logic.

Most mathematical statements are combinations of simpler statement formed through some choice of the words ”not”,”and”,”or”,”if ...then” and “if and only if”. These are called logical connectives or simply connectives and are denoted by the following symbols:

Connective Symbol Formal name

Not ∼ or ¬ Negation

And ∧ Conjunction

Or ∨ Disjunction

If...then → Conditional

If and only if ↔ Biconditional

Table 2.1 Symbols of connectives

2.5.1 NegationDefinition of negationIf P is a statement variable,the negation of P is “not P” or it is not the case that P” and is denoted by ∼ P.It has opposite truth value from P.

The negation statement is generally formed by introducing the word “not” at a proper place in statement with the phrase “It is not the case that” and read as “not P”.

Let P be a statement .The negation of P, written ∼ P or P¬ is the statement obtained by negating statement P.

If the truth value of P is true then truth value of ∼ P is false, and if the truth value of P is false then truth value of ∼ P is true.

Thisdefinitionofnegationissummarizedbythetruthtablebelow.

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P ∼ P

T F

F T

Table 2.2 Truth Table of Negation

ExampleP:The integer 10 is evenThen ∼ P: The integer 10 is not even

P:London is a city∼ P:It is not case that London is a city∼ P:London is not a city

P:I went to my class yesterday.∼ P:I did not go to my class yesterday∼ P:I was absent from my class yesterday.

P:I went to my class yesterday.∼ P:I did not go to my class yesterday∼ P:I was absent from my class yesterday

Negationiscalledconnectivesalthoughitonlymodifiesastatementoravariable.

2.5.2 ConjunctionLet P and Q be statements. The conjunction of P and Q, written P ∧ Q,is the statement formed by joining statements P and Q using the word “and”. The statement P∧Q is true if both P and Q are true; otherwise P∧Q is false.

The symbol ∧ is called “and”. Let P and Q be statements. The truth table of P∧ Q is given below.

DefinitionIf p and q are statement variables, the conjunction of p and q is “p and q”, denoted p∧q.The compound statement p∧q .The compound statement p ∧ q is true when both p and q are true; otherwise, it is false.

P Q P∧Q

T T T

T F F

F T F

F F F

Table 2.3 Truth table of conjunction

ExampleP: 2 is an even integer,Q: 7 divide 14R:2 is an even integer and 7 divides 14.

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P:It is raining todayQ:There are 20 tables in this room.R:It is raining today and there are 20 tables in this room.

Jack and Jill went up the hill.From this statement we get two statement Jack went up the hill and Jill went up the hill.Then the given statement can be written symbolic from P∧Q.

2.5.3 Disjunction Let P and Q be statements. The disjunction of P and Q ,written P ∨ Q ,is statement formed by putting statements P and Q together using the word “Or”. The truth value of the statement P∨Q is T if atleast one of statements P and Q is true. The symbol ∨ is called “Or”, for the statement P ∨ Q is given below.

DefinitionIf P and Q are statement variables, the disjunction of P and Q is “P or Q”, denoted P ∨ Q.The compound statement P ∨ Q is true if atleast one of P or Q is true; it is false when both P and Q are false.

P Q P∨ Q

T T T

T F T

F T T

F F F

Table 2.4 Truth table for disjunction

ExampleP:22+33 is an even integerQ:22+33 is an odd integer then P ∨ Q:22+33 is an even integer or 22+33 is an odd integer ORP ∨ Q :22+33 is an even integer or an odd integer The notation for inequalities involves “and” and “or” statements.Let a,b and c be particular real numbers.a≤ b means a < b or a= ba<b<c means a<b and b<c∼ is a unary operation while ∨ and ∧ are binary operations.3 0r -5 is negative –truth value is false

or π is an integer ---truth value is false

2.5.4 Conditional or ImplicationLet P and Q be two statements.Then “If P,then Q” is the statement called an Implication or conditional statement,written P → Q .

The statement P → Q has a truth value F when the truth value of P is true and Q is false.Otherwisetruth value of conditional or implication is T.The statement P is called the antecedent or hypothesis and q is called consequent or conclusion in P→Q.Accordingtothedefinition,itisnotnecessarythattherebeanykindofrelationbetweenPandQ in order to form P → Q.The statement P → Q is also to be read asP implies QOrPissufficientforQ

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OrQ if POrQ whenever P.In the implication P → Q,P is called the hypothesis and Q is called the conclusion. The truth table of P → Q is given below.DefinitionIf P and Q are statements,the statemen”if P then Q” or “P implies Q”,denoted P→Q is called the conditional statement,or implication.

P Q P → Q

T T T

T F F

F T T

F F T

Table 2.5 Truth table for implication

ExampleIf today is Sunday,then i will go for walk.Let P:Today is SundayQ:I will go for walk

Variety of terminaology:If P then Q Q if PP implies Q Q when PP only if Q Q follows from PPissufficientforQQisnecessaryforP

2.5.5 Biconditional or BiimplicationLet P and Q be two statements.Then “P if and only if Q”,written P↔Q is called the Biimplication or biconditional of the statement P and Q.The statement P↔Qmayalsobereadas“PisnecessaryandsufficientforQ”or“QisnecessaryandsufficientforP”or“QifandonlyifP”or“QwhenandonlywhenP”.WedefinethattheBiimplicationP↔Q is considered to be true when both P and Q have the same truth values and false otherwise. It is denoted by P↔Q.The truth table is given below.

P Q P↔Q

T T T

T F F

F T F

F F F

Table 2.6 Truth table of biimplication

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2.5.6 Contrapositive, Converse and InverseThe contrapostive of P→ Q is ∼Q→∼PThe converse of P→ Q is Q→PThe inverse of P→ Q is ∼P → ∼Q

2.6 Tautology ,Contradiction and ContingencyA tautology is a statement form where its truth values in all rows in the truth table are always true. A contradiction is a statement form where its truth values in all rows in the truth table are always false.

A contingency is a statement form that is neither tautology nor contradiction. Normally t is denoted to use tautology and c is used to denote a contradiction.

ExampleLet P, Q, R be statement variables. Show that the statement form∼P ∨ P is a tautology∼P ∧ P is a contradiction(P∧Q) ∨ ∼ R is a contingency

a.

P ∼P ∼P ∨ P

F T T

T F T

Table 2.7 ∼P ∨ P is a tautology

b.

P ∼P ∼P ∧ P

F T F

T F F

Table 2.8 contradiction

c.

P Q R P∧Q ∼R (P ∧ Q) V ∼R

F F F F T T

F F T F F F

F T F F T T

F T T F F F

T F F F T T

T F T F F F

T T F T T T

T T T T F T

Table 2.9 contingency

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2.7 Laws of AlgebraFollowing are the laws of algebra2.7.1 Identity LawP ∨ P ≡ PP ∨ T≡ TP ∨ F ≡ PP∧ P ≡ PP∧ T ≡ PP∧ F ≡ FT→P≡ PF→P≡ TP→P≡ TP→T≡ TP→F≡ ∼PP↔ P≡ TP↔ T≡ PP↔ F≡ ∼P

2.7.2 Commutative LawP∨Q ≡ Q ∨PP∧Q≡Q∧PP→Q ≠ Q→ PP↔ Q = Q↔P

2.7.3 Complement LawP ∨ ∼ P ≡TP ∧ ∼ P ≡ FP→∼ P≡∼ PP↔∼ P≡F∼ P→P≡P

2.7.4 Double Negation∼(∼ P) ≡ P

2.7.5 Associative LawP∨(Q∨R)≡ (P∨Q)∨RP∧(Q∧R)≡(P∧Q)∧R

2.7.6 Distributive LawP∨(Q∧R)≡ (P∨Q)∧(P∨R)P∧(Q∨R)≡ (P∧Q)∨(P∧R)

2.7.7 Absorption LawP∨(P∧Q) ≡PP∧(P∨Q) ≡P

2.7.8 Demorgan’s Law∼(P∨Q)≡∼P ∧∼Q∼(P∧Q)≡∼P ∨∼Q

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2.7.9 Equivalance of ContrapositiveP→Q≡∼Q→∼P

2.7.10 OthersP→Q≡∼P ∨QP↔Q≡(P→Q)∧(Q→P)

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SummaryMathematical Logic is a tool for providing a precise meaning to mathematical statements.•A statement is a declarative sentence which is either true or false but not both.•The truth value of a proposition is true (T) if it is of true proposition and false (F) if it is false proposition.•A table that gives the truth value of the compound statement in terms of its component part is called a truth •table.A compound statement is a combination of two or more statements.•All declarative sentences to which it is possible to assign one and only of the two possible truth values are •called statements.It is possible to construct rather complicated statements from simpler statements by using certain connecting •words or expressions known as “sentential connectives”.The negation statement is generally formed by introducing the word “not” at a proper place in statement with •the phrase “It is not the case that” and read as “not P”.Let P and Q be two statements. Then “If P, then Q” is the statement called an Implication or conditional statement, •written P → Q.

ReferencesFulda, J. S., 1993. • Exclusive Disjunction and the Bi-conditional: An Even-Odd Relationship, Mathematics Magazine.Hallie, P. P., 1954. • A Note on Logical Connectives, Mind 63.Bartlett, A., • Simple Mathematical Logic, [Video Online] Available at: <http://www.youtube.com/watch?v=PlSfS5-jU_g > [Accessed 31 August 2012].Dr. Kirthivasan, K., • Propositional Logic, [Video Online] Available at: <http://www.youtube.com/watch?v=xlUFkMKSB3Y > [Accessed 31 August 2012].Lifschitz, V., • Lecture notes on mathematical logic, [pdf] Available at: <http://www.cs.utexas.edu/~vl/teaching/388Lnotes.pdf > [Accessed 31 August 2012].Simpson, S., • Mathematical Logic, [Online] Available at: <http://www.math.psu.edu/simpson/courses/math557/ > [Accessed 31 August 2012].

Recommended ReadingDean McCullough, P., 1971. • Logical Connectives for Intuitionist Propositional Logic, Journal of Symbolic Logic.Wansing, H., 2006. • Logical Connectives for Constructive Modal Logic.Hallie, P. P., 1954. • A Note on Logical Connectives, Mind 63.

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Self Assessment:“Canada is a country” .what kind of statement is it?1.

Primitive statementa. Compound statementb. Elementary statementc. Primary statementd.

“A dog or a car is an animal”. What kind of statement it is?2. Primitive statementa. Compound statementb. Elementary statementc. Primary statementd.

What is the symbol for negation?3. a. ∧b. ∨c. ∼d. →

What is the symbol for conjunction?4. a. ∧b. ∨c. ∼d. →

What is the symbol for disjunction?5. a. ∧b. ∨c. ∼d. →

What is the symbol for implication?6. a. ∧b. ∨c. ∼d. →

If P is true and Q is False what is the value of P 7. ∨ Q?Ta. Fb. Invalidc. no valued.

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If P is true and Q is False what is the value of P 8. ∧ Q?Ta. Fb. Invalidc. no valued.

If P is true and Q is False what is the value of P 9. → Q?Ta. Fb. Invalidc. no valued.

If P is true and Q is False what is the value of P 10. ↔ Q?Ta. Fb. Invalidc. no valued.

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Chapter III

Set Theory

Aim

The aim of this chapter is to:

defineaset•

discuss different types of set•

elucidate operations on sets•

Objective

The objectives of this chapter are to:

discuss the types of sets•

explicate null and universal set•

explain Demorgan’s Law•

Learning outcome

At the end of this chapter, you will be able to:

understand standard set•

comprehend the concept of intersection and disjoint sets•

understand operati• ons on sets.

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3.1 Definition of a SetAsetisacollectionofwelldefinedobjectenclosedwithincurlybrackets,generallydenotedbyuppercaseletters.

The objects which form the set are called the elements or members of the set. The elements are denoted by lower case letters.

ExampleA= 1, 2, 3... 100 represents set of numbers from 1 to 100V= a, e, i, o, u represents set of vowels in English alphabetsThe following notation is used to show the set membershipx ∈ A means that x is a member of the set A.x ∉ A means that x is not a member of the set A.

3.2 Standard SetsFollowing are the standard setsN= set of natural numbers = 1, 2, 3, 4...Z= set of all integers = ...,-3,-2,-1, 0,1,2,3...R= set of all rational numbers = p/q: p is integer, q≠0Q=set of all real numbers

3.3 Representation of setFollowing are the method for representing sets3.3.1 Tabular Form/Roaster MethodIn the tabular form, all elements of the set are enumerated or listed.

ExampleThe set of natural numbers from 1 to 100 is given byN= 1, 2, 3, 4...100The set of vowels is given byV= a, e, i, o, u

3.3.2 Rule MethodUnderthismethod,thedefiningpropertyofthesetisspecified.IfalltheelementsinthesethaveaPropertyP,thenwecandefinethesetasA=x: x has the property P

ExampleN=x: 1≤x≤100, x is a natural number or x/1≤x≤100,x∈NWhere, N represents natural numberV=x/x is a vowel of English alphabet

3.3.3 Descriptive FormIn this method, the elements in the set are verbally described within the curly brackets

Example:N= natural numbers from 1 to 100V= vowels of English alphabet

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3.4 Types of SetsFollowing are the types of sets

3.4.1 Finite SetAsetwithafiniteorcountablenumberofelementsiscalledafiniteset.Example:V= a, e, i, o, u

3.4.2 Empty or Null SetA set which contains no elements is called an empty set or a null set. An empty set is denoted as or φ.Example:A= A= φ

3.4.3 SubsetSet P is a subset of a set Q, symbolised by P⊆ Q, if and only if all the elements of set P are also the element of set Q.

Note:every set is a subset of itself•null set is a subset of every set•

Example:Let A = 1, 2, 3Then , 1, 2, 3,1,2,1,3,2,3.1,2,3 are the subsets of the set A=1,2,3.

3.4.3.1 Proper SubsetSet G is a proper subset of H, symbolised by G ⊂ H, if and only if all the elements of set G is elements of set H and set G ≠ set H.

That is set H must contain at least one element not in set G.ExampleConsider above example, ,1,2,3,1,2,1,3,2,3 are the proper subset of A=1,2,3.

3.4.3.2 Improper SubsetSet S is an improper subset of T, symbolised by S ⊆ T , if and only if all the elements of set S are the elements of set T and set S = set T.ExampleIn the above example 1,2,3 is an improper subset of set A=1,2,3

3.4.4 Infinite SetNeitheranemptysetnorafinitesetiscalledaninfiniteset.Suchsetwillcontaininfinitelymanyelements.ExampleN= 1, 2, 3...Z= ...-3,-2,-1, 0,1,2,3...

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3.4.5 Disjoint SetsTwo sets are said to be disjoint sets if they do not have any common elements.Example IfA=x: x is an even numberB=y: y is an odd numberThen A and B will be disjoint sets.

3.4.6 Overlapping SetsTwo sets are said to be overlapping sets, if they have some common elements.ExampleA= 1, 2, 3, 4, 5B= 3, 4, 5, 6, 7

These sets are overlapping sets as the elements 3, 4 and 5 are common to both set.The element common to both sets is called the intersection of the two sets and is denoted as A ∩ B.On the other hand, the addition of two sets is called the union of two sets and is denoted by A ∪ B.

3.4.7 Universal SetA universal set is the set that contains the element of all the sets under consideration. It is usually denoted by U, S or Ω.ExampleA=4,7,8,9, B=-4,-2,0,1,4,7,10.The set of Integers, I= ...,-2,-1, 0, 1, 2... will be the universal set for A and B.

3.4.8 Equal SetTwo sets are said to e equal if they contain the same elements.

ExampleA= 2, 4, 6, 8 B=x: x is an even number between 1 and 9 are equal set as they contain the same elementsP= 1, 3, 6 Q= 6, 1, 3 is also an equal set.

3.4.9 Complement SetGiven a set A, the complement set is the set that contains elements not belonging to A and is denoted by A’. The union (will be discussed in operations on set) of the given set and its complement will give the universal set. That is A ∪ A’ = U.

ExampleA=x: x is a Mathematics book in the LibraryA’ =y: y is not a Mathematics book in the LibraryAnd the universal set in this case is U= the set of all books in the Library

3.4.10 Equivalent SetTwo sets are said to be equivalent set if number of elements in one set is equal to the number of elements in the other set. The number of elements in a set is known as Cardinal number of the set.

So, if the cardinal numbers of two sets are equal then those two sets are said to be equivalent set. Cardinal number of the set A is represented by n (A).ExampleA= 1, 2, 3, 4 B= a, b, c, d

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Here, n (A) =4 and n (B) =4So the sets A and B are said to be Equivalent sets.

3.5 Illustration of Various SetsLet us consider three sets A, B and C in the rule form as A= x: x are the vowels, B= x: x are the consonants C=x:xisthefirst10alphabetsofEnglish

In tabular formA=a,e,i,o,u, B=b,c,d,f,g,h,j,k,l,m,n,p,q,r,s,t,v,w,x,y,z,C=a,b,c,d,e,f,g,h,i,j

set A and B are disjoint sets as they no elements of A is present in B and vice versa.•set A and C are overlapping sets as both contains the element a, e, i in common.•set B and C are also overlapping set as both contain the element b, c, d, f, g, h, j in common.•set of all alphabets of English is the universal set for A, B, C. U= a, b, c, d... z•set A is a complement of set B as it contains elements not contained in set A and union of set A and set B will •give the Universal set U.

3.6 Basic Operations on SetsFollowing are the basic operations on sets

3.6.1 Intersection of Two SetsThe intersection of two sets is the set of elements common to both the given sets. The intersection of two sets A and B is denoted as A ∩ B.

Innotationform,wecandefinetheintersectionoftwosetsAandBasA ∩ B = x: x ∈ A, x ∈ B.

If, A ∩ B = φ, then A and B are said to be disjoint sets. If A ∩ B ≠ φ,then A and B are called overlapping sets.

ExampleGivenA= 1, 2, 3, 4, 5B= 4, 5, 6, 7, 8Then A ∩ B = 4, 5

3.6.2 Union of Two SetsThe union of the two sets is the set containing the elements belonging to A and also the elements belonging to B.The union of these sets is denoted as A ∪B.Innotationform,wecandefinetheunionofthetwosetsasA∪ B = x: x ∈ A, x ∈ B, x ∈ A ∩ B.

ExampleA= 1, 2, 3, 4, 5,B= 4, 5, 6, 7, 8Therefore, A ∪ B = 1, 2, 3, 4, 5, 6, 7, 8

3.6.3 Relative Complement or Difference of Two SetsThe difference of two sets A and B is the set of elements that belong to A but do not belong to B.The difference of two sets is denoted by A–B.Innotationform,wecandefinethedifferenceoftwosetsasA – B = x: x ∈ A, x ∉ BSimilarly, B – A =x: x ∈ B, x ∉ A

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ExampleA= 1, 2, 3, 4, 5B= 4, 5, 6, 7, 8Then, A – B = 1,2,3 Though 4,5 belongs to A it also belongs to B so that is not included .

3.6.4 Complement of a SetComplement set is the set that contains elements not belonging to A but belonging to the universal set. In notation form,wecandefinethecomplementofasetasA’=x:x∉ A, x ∈ U.

ExampleGiven U= 1,2,3,4,5,6,7,8,9,10A= 1, 2, 3Therefore A’ = 4, 5, 6, 7, 8, 9, 10

3.6.5 Symmetric Difference of Two SetsThe symmetric difference of two sets is the union of both the relative complements.A ∆ B = (A – B) ∪ (B – A)

It can also be expressed as relative complement of union of the two sets and intersection of the two sets.A ∆ B = (A ∪ B) – (A ∩ B)

ExampleGiven A = 1, 2, 3, 4, 5

B= 4, 5, 6, 7, 8

Solution 1:A ∆ B = (A – B) ∪ (B – A) A – B = 1, 2, 3B – A = 6, 7, 8So, A ∆ B = 1, 2, 3 ∪ 6, 7, 8Therefore A ∆ B = 1, 2, 3, 6, 7, 8

Solution 2:A ∆ B = (A ∪ B) – (A ∩ B) A ∪ B = 1, 2, 3, 4, 5, 6, 7, 8A ∩ B = 4, 5A ∆ B = 1,2,3,4,5,6,7,8 - 4,5Therefore A ∆ B = 1, 2, 3, 6, 7, 8

3.7 Properties of SetFollowing are the properties of sets

3.7.1 Commutative LawA ∪ B = B ∪ AA ∩ B = B ∩ AA ∆ B = B ∆ A

3.7.2 Associative LawA ∩ (B ∩ C) = (A ∩ B) ∩ CA ∪ (B ∪ C) = (A ∪ B) ∪ CA ∆ (B ∆ C) = (A ∆ B) ∆ C

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3.7.3 Distributive LawA ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

3.7.4 Identity LawA ∪ φ = AA ∩ U = Aφ’ = UU’ = φ

3.7.5 Complement LawA ∪ A’ = UA ∩ A’ = φ

3.7.6 Idempotent LawA ∪ A = AA ∩ A = A

3.7.7 Bound LawA ∪ U = UA ∩ φ = φ

3.7.8 Absorption LawA ∪ (A ∩ B) = AA ∩ (A ∪ B) = A

3.7.9 Involution Law(A’)’= A

3.7.10 De Morgan’s LawThe complement of the union of two sets is equal to the intersection of their complements.(A ∪ B) ’ = A’ ∩ B’

The complement of the intersection of two sets is equal to the union of their complements.(A ∩ B)’ = A’ ∪ B’A - (B ∩ C) = (A – B) ∪ (A – C)A – (B ∪ C) = (A - B) ∩ (A – C)

3.7.11More Resultsn (A ∪ B) = n (A) + n (B) – n (A ∩ B)n(A ∪ B ∪ C) = n(A) +n(B) +n(C)- n( A∩ B)-n(B ∩ C) –n(A ∩ C) + n( A∩B∩C)

Solved ExamplesExample 1If A= 1,2,3,4 , B= 2,4,6,8 ,C= 3,4,5,6,8, U =1,2,3,4,5,...,10Find A’, B ∪ C, A ∩ C

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Solution:A’ = 5, 6, 7, 8, 9, 10B ∪ C = 2, 3, 4, 5, 6, 8A ∩ C = 3, 4

Example 2If A= 1,2,3,4 , B= 2,4,6,8 ,C= 3,4,5,6,8, U =1,2,3,4,5,...,10Find A – B, B – C, (A ∪ B)’

Solution: A – B= 1, 3B – C= 2A ∪ B = 1, 2, 3, 4, 5, 6, 8(A ∪ B)’= 7, 9, 10

Example 3If U =1,2,3...10 A=3,4,6,10 B=1,2,4,5,6,8 Verify De Morgan’s laws of complementation.(A ∪ B)’ = A’ ∩ B’

Solution: LHS: (A ∪ B)’A ∪ B = 1, 2, 3, 4, 5, 6, 8, 10(A ∪ B)’ = 7, 9 --------- (1)RHS: A’ ∩ B’A’ = 1, 2, 5, 7, 8, 9B’ = 3, 7, 9, 10A’ ∩ B’ = 7, 9 --------- (2)From (1) and (2), it is clear (A ∪ B)’ = A’ ∩ B’

(A 1. ∩ B)’ = A’ ∪ B’Solution: LHS = (A ∩ B)’A ∩ B = 4, 6(A ∩ B)’ = 1,2,3,5,7,8,9,10-----------(1)Now, A’ = 1, 2, 5, 7, 8, 9B’ = 3, 7, 9, 10Therefore, A’ ∪ B’ = 1,2,3,5,7,8,9,10-------(2)From (1) and (2) it is clear that (A ∩ B)’= A’ ∪ B’HenceVerified.

Example 4If A= -8,-7,-5, 1, 2, 4B= -7, 1, 3, 4, 5, 6C= -8,-5, 2, 4, 6, 7Verify A- (B ∪ C) = (A – B) ∩ (A – C)

Solution: LHS: A – (B ∪ C) B ∪ C = -7, 1, 3, 4, 5, 6,-8,-5, 2, 7

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Therefore A – (B ∪ C) = ------ (1)RHS: (A – B) ∩ (A – C)Now, A – B = -8,-5, 2A – C = 1,-7(A – B) ∩ (A – C) = --------- (2)

From (1) and (2), it is clear that A – (B ∪ C) = (A – B) ∩ (A – C)Henceverified.

Example 5If A= -9,-7,-6,-3,0,2,B= -7,-3,0,4,5,6 C=-9,-6,2,-7,8,Verify A – (B ∩ C) = (A – B) ∪ (A – C)

Solution:L H S: A – (B ∩ C)B ∩ C = -7

Therefore A – (B ∩ C) = -9,-6,-3, 0, 2 -------- (1)Now, A – B = -9,-6, 2A – C = -3, 0RHS: (A – B) ∪ (A – C) = -9,-6, 2,-3, 0 ----- (2)

From (1) and (2), it is clear A – (B ∩ C) = (A – B) ∪ (A – C)Henceverified.

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SummaryAsetisacollectionofwelldefinedobjectenclosedwithincurlybrackets.•The objects which form the set are called the elements or members of the set.•In the tabular form, all elements of the set are enumerated or listed.•Set P is a subset of a set Q, symbolised by P• ⊆ Q, if and only if all the elements of set P are also the element of set Q.A set which contains no elements is called an empty set or a null set.•Two sets are said to be disjoint sets if they do not have any common elements.•Two sets are said to be overlapping sets, if they have some common elements.•A universal set is the set that contains the element of all the sets under consideration.•Two sets are said to be equal if they contain the same elements.•The intersection of two sets is the set of elements common to both the given sets.•The union of the two sets is the set containing the elements belonging to A and also the elements belonging to •B.A • ∆ B = (A – B) ∪ (B – A) Demorgan’s Law(A • ∪ B) ’ = A’ ∩ B’

ReferencesAkekar, R., 2008.• Discrete Mathematics: Set theory, 2nd ed., Dorling Kindersley Publication India.T. • Veeraranjan, 2008. Discrete Mathematics with graph theory and Combinatorics: Set theory, 7th ed., McGraw- Hill Publication.Lipchutz, S., 1998.• Set Theory and related topics, 2nd ed., McGraw-Hill Publication.Arithmetic and geometric progressions• , [Online] Available at: < http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-apgp-2009-1.pdf > [Accessed 31 August 2012].Arithmetic and geometric progressions,• [Online] Available at: < http://maths.mq.edu.au/numeracy/web_mums/module3/Worksheet36/module3.pdf > [Accessed 31 August 2012].2011, • Arithmetic & Geometric progressions, [Video Online] Available t: < http://www.youtube.com/watch?v=ze0hNuxJaVE > [Accessed31 August 2012].Lee, D., • Arithmetic Progression, [Video Online] Available at: < http://www.youtube.com/watch?v=kGh9BtU-OVU > [Accessed 31 August 2012].

Recommended Reading Waters, D., 2006. • Quantitative Methods for business, Set Theory, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative Methods, Set theory, Elsevier.Slater, J, C., 2007. • Quantitative Methods, Set theory, Thomson Learning.

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Self AssessmentA____isacollectionofwelldefinedobjectenclosedwithincurlybrackets.1.

Seta. Subsetb. Objectc. Elementd.

A set which contains no elements is called _____ set.2. equala. equivalentb. emptyc. finited.

Two sets are said to be __________ if they do not have any common elements.3. disjoint setsa. overlapping setsb. intersection setsc. equal setsd.

Two sets are said to be ______ if number of elements in one set is equal to the number of elements in the other 4. set.

disjoint seta. equivalent setb. overlapping setc. intersection setd.

If A = 1,2,3,4,5 and B=4,5,6,7,8 ,what is A – B ?5. 4,5a. 6,7,8b. 1,2,3c. 1,2,3,4,5d.

Is P= 1, 3, 6 Q= 6, 1, 3 an equal set?6. Truea. Falseb.

Is A= 1, 2, 3, 4 and B= a, b, c, d an equivalent set?7. Truea. Falseb.

If A= 1,2,3,4 and U =1,2,3,4,5,...,10,What is the value of A’?8. 1,2,3,4,5a. 4,5,6,7b. 7,8,9,10c. 5,6,7,8,9,10d.

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If B= 2,4,6,8 and C= 3,4,5,6,8 what is the value of B 9. ∪ C?2,3,4,5,6,8a. 2,4,6,8b. 3,4,5,6,8c. 4,6,8d.

If A= 1,2,3,4 and C= 3,4,5,6,8 what is the value of A 10. ∩ C?1,2,3,4a. 3,4b. 3,4,5,6,8c. 1,2,3,4d.

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Chapter IV

Progression

Aim

The aim of this chapter is to:

introduce the concept of arithmetic progression•

definetheconceptofgeometricprogression•

explicate the formulae for arithmetic progression•

Objective

The objectives of this chapter are to:

elucidate progression•

discuss arithmetic and geometric progression•

explain formulae for arithmetic and geometric progression•

Learning outcome

At the end of this chapter, you will be able to:

understand the concept of arithmetic and geometric progression•

identify the arithmetic representation of a series•

comprehend the geom• etric representation of series

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4.1 IntroductionProgressionisapatternofnumberswhereinaquantityincreasesordecreasesprogressivelybyfixedamount.Inotherwords,consecutivetermintheseriesincreaseordecreasebyafixednumber.Asuccessionofnumbersformedandarrangedinadefiniteorderaccordingtocertaindefiniterule,iscalledaprogression.

4.2 Arithmetic ProgressionIf quantities increase or decrease by common difference, then they are said to be in arithmetical progression (AP).If each term of a progression differs from its preceding term by a constant, then such a progression is called arithmetical progression. This constant difference is called the common difference of the AP.

Let us look at few lists as follows3,5,7,9,11,13,15…•4,8,12,16,20,24…•1, 4, 7,10,13,16 …•-10,-6,-2, 2,6,10…•20, 18, 16, 12, 10….•

Above we notice that the numbers in each of the list are different but there is a pattern that helps us determine the next so let us identify the patterns in the above list

leavingthefirstnumbereachsuccessivenumberisobtainedbyadding2•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding4•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding3•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding4•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding-2•

Thus we observe that each list starts with a number chosen arbitrarily•everysuccessivenumberinthelistisformedbyaddingthesamefixednumbertothepreviousnumber•

Thus to generalize it, we haveLet a be any starting termd be the common difference between two successive terms

Then a, a+d,(a+d) +d,((a+d)+d)+d,…

Therefore, a, a+d, a+2d, a+3d… are the successive terms of an arithmetic progression.

The points to be remembered are:The sequence of the next number is very important and we cannot change the sequence as it will destroy the •patternWhilefindingthecommondifferencewemustsubtractthenthtermfrom(n+1)thtermevenifitissmaller•

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4.3 Formulae for Arithmetic Progression

4.3.1 The general form of an AP The general form of an AP is given bya,a+d,a+2d,a+3d … a(n-1)d

Where,aisthefirstterma+ (n-1) d is the nth termd is common difference

4.3.2 The nth term of an APThe nth term of an AP is given by

= a + (n-1) d

4.3.3 Sum of first n terms ( ) of an AP TheSumoffirstnterms( ) of an AP is usually denoted by and is given by

= (a + l) = (a1+ an)Or

= [a + (a+ (n-1) d)] = (2a+ (n-1) d)

Where,aisthefirsttermofAPd is the common difference l is the last term an is the nth term

4.4 Arithmetic MeanWhen three quantities are in AP, the middle one is said to be the arithmetic mean of the other two.If a and b are the two quantities in AP and if A is the arithmetic mean then,A = (a+b)/2a, A,b are in AP.

If a and b are two given quantities in A.P., and if n arithmetic means are to be inserted between a and b ,means a,A1,A2,A3,.......,An,b are in AP where A1,A2,A3,.......,An are means.Then‘m’ th arithmetic mean (Am) is given by a +m(b-a)/(n+1)

WhereaisthefirsttermoftheAPb is the last term.n is the number of arithmetic means between themm is the mth arithmetic mean

4.5 Geometric ProgressionIf quantities increase or decrease by common multiple, then they are said to be in geometric Progression (GP)Let us look at a few lists as follows:

3, 6, 12, 24, 48, 96, 192…, •8, 4, 2, 1, 1/2, 1/4, 1/8…•

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1,3,9,27,81,243,729 …•3,-6, 12,-24, 48,-96, 192…•6561,-2187, 729,-243, 81,-9…•

Above we notice that the numbers in each of the lists are different but there is a pattern that helps us determine the next term so let us identify the patterns in the above lists

leavingthefirstnumbereachsuccessivenumberisobtainedbymultiplyingby2•leavingthefirstnumbereachsuccessivenumberisobtainedbymultiplyingby½ordividingby2•leavingthefirstnumbereachsuccessivenumberisobtainedbymultiplyingby3•leavingthefirstnumbereachsuccessivenumberisobtainedbymultiplyingby-2•leavingthefirstnumbereachsuccessivenumberisobtainedbymultiplyingby-1/3ordividingby-3•

Thus we observe that each list starts with a number chosen arbitrarily•everysuccessivenumberinthelistisformedbymultiplyingordividingwiththesamefixednumbertothe•previous numberTherefore, to generalize what we have Let a be any starting termr be the common multiple between two successive termsThen, a, ar, (ar) r, ((ar) r) r…Therefore a, ar, ar2, ar3… are the successive terms of a geometric progression.The points to remember the sequence of the next number is very important and we cannot change the sequence as it will destroy the •patternwhilefindingthecommonmultiplebydividing(n+1)thtermevenifitsmaller•

4.6 Formulae for Geometric Progression

4.6.1 The general form of a GPThe general form of a GP is a, ar, ar2, ar3, …. arn-1

Whereaisthefirstterma(r) n-1 is the nth term r is the common multiple

4.6.2 The nth term Tn of a GPThe nth term Tn of a GP is given by Tn = a(r) n-1

4.6.3 The sum of first n terms Sn of a GPThesumoffirstntermsofaGPisusuallydenotedbySn and is given by

Sn = for r>1

Sn = for r<1

Sn = forr<1andnisinfinite

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WhereaisthefirsttermofAPr is the common ratio or multiplel is the last terman is the nth term.

4.7 Geometric MeanWhen three quantities are in GP, the middle one is said to be the geometric mean of the other two.

If a and b are two quantities in GP and if G is the geometric mean then,G= A, G, b are in GP.

If a and b are two given quantities in G.P., and if n geometric means are to be inserted between a and b, means a, G1, G2, G3... Gn, b are in GP where, G1, G2, G3... Gn, are means

Then‘m’ th geometric mean(Gm) is given by WhereaisthefirsttermoftheAP b is the last termn is the number of geometric means between themm is the mth geometric mean

4.8 Harmonic ProgressionIf quantities increase or decrease such that the inverse of the quantities have a common difference (are in a arithmetic progression),then they are said to be in Harmonic Progression(HP).

Let us look at a few lists as follows:10, 12, 15...•3, 3/2, 1...•10, 20/3, 5, 4...•

If we look at the inverse of the following is the format of the lists:1/10, 1/12, 1/15•1/3, 2/3, 1....•1/10, 3/20, 1/5, 1/4•

Above we notice that the inverse of the given numbers in each of the lists are different but there is a pattern which isanAP.Letusfindthecommondifference

leavingthefirstnumbereachsuccessivenumberisobtainedbyadding-1/60•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding1/3•leavingthefirstnumbereachsuccessivenumberisobtainedbyadding1/20•

Thus we observe thateach lists starts with a number chosen arbitrarily•all the inverse of the numbers are in AP•

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Thus to generalize what we haveLeta be any starting termd be the common difference between two successive terms

Then,a, a+d,(a+d)+d,((a+d)+d)+d,...Therefore a, a+d, a+2d, a+3d... are the successive terms of an arithmetic progressionTherefore 1/a, 1/a+d, 1/a+2d, 1/a+3d ... will be the harmonic progression.

The points to remember are:the sequence of the next number is very important and we cannot change the sequence as it will destroy the •patternwhilefindingthecommondifferencewemustsubtractthen• th term from (n+1) th term even if it is smaller

4.9 Formulae for Harmonic Progression4.9.1 The General Form of HPThe general form of a HP is:1/a, 1/a+d, 1/a+2d, 1/a+3d...1/a+ (n-1) dWhere a is the 1st term1/a+ (n-1) is the nth termd is the common difference

4.9.2 The nth term (Tn)of a HPThe nth term (Tn) of a HP is given by

Tn =

4.10 Harmonic meanWhen three quantities are in HP,the middle one is said to be the Harmonic mean of the other two. If a and b are two quantities in HP and if H is the Harmonic mean then,

H=

a, H, b are in HP.

4.11 Comparison between AP and GPFollowing is the difference between AP and GP

Description Arithmetic Progression Geometric Progression

Principal Characteristic Common Difference(d) Common Ratio(r)

nth term Tn=a+(n-1)d Tn = a r (n-1)

Mean A= (a+b)/2 G=(ab)1/2

Sum of First n terms Sn=n/2[2a+(n-1)d]=n/2[a+l] Sn=a(1-rn)/(1-r)

‘m’ th mean a+[m(b-a)/(n+1)] A(b/a) m/(n+1)

Table 4.1 Comparison between AP and GP

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4.12 Important Rules on Arithmetic mean(AM),Geometric Mean (GM) and Harmonic Mean(HM)

AM • ≥ GM ≥ HMGM = • (AM x HM)

Solved Examples

Example 1Find the sum of the series 5/2,4,11/2,7,….upto 21 terms

Solution:d=4-5/2 = 3/2; n=21; a=5/2 (from the data given)Sn = [21/2] x [2 x (5/2) +20 x (3/2)] = [21/2] x [5+30] = 735/2 =367.50

Example 2Find the 74th term of the series 3, 7, 11, 15…

Solution:a=3; d=7-3=4; n= 74Tn = a + (n-1) dT74 = 3+73 x 4 =3 + 292=295

Example 3Find the total earning over 6 years for a person whose salary is Rs.30000 p.a and increases by 1500 p.a, every six months.

Solution:For each six months he earns half his annual salaryTherefore, a =30000/2=15000Tofindthesumof12termsd=750S12 =12/2[2 x (15000) +11 x 750] = 6 [30000+8250]= Rs.229500

Example 4Findthesumoffirst6termsoftheseries3,9,27,81….n=6; r= 27/9 =3; a=3Sn = 3(1-36 ) / (1-3) =1092

Foraninfiniteseries,S∞ → ± ∞ (if |r| ≥ 1)S∞ = a/ (1-r) (if |r| ≤ 1)

Example 5Findthesumoftheinfiniteseries1,0.5,0.25…

Solution:r=0.5 since -1≤ 0.5 ≤ 1 S =a/ (1-r)S= 1/ (1 - 0.5) =1/0.5 = 2

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Example 6Find the 10th term of the series 4, 16,64,256,1024…

Solution:The given series is in geometric progression where a=4 and r=4TofindT10T10= a r (10-1) =4 x 49= 410

Example 7Thesumofntermsoftheseries2,5,8…is950findn

Solution:Given Sn = 950; a=2 and d=5-2=3Therefore Sn = a + (n-1) d950 = 2 + (n-1) 3950-2 = (n-1) 3948= (n-1) 3n-1 = 948/3n-1 = 316n=316+1=317

Therefore the value of n is 317.

Example 8SumofaGPwhosecommonratiois3is728.Thelasttermis486.Findthefirstterm.

Solution: Sum= Sn =728; r= 3; l=486LetthefirsttermbeaandnumberoftermsbenBy the formula,Tn=a rn-1

486 = a 3 n-1

Therefore,3n /3 =486/aWhich implies 3n=1458/aAlso the sum of n terms = Sn=a (3n-1)/ (3-1) =728Substituting the value of (3n) in the above equationFrom above equation we get, a [1458/a-1] =1456Therefore, a [1458 – a] = 1456.aa2=2.a Which implies a=2

Example 9Find the geometric mean between the terms 4 and 25

Solution: The geometric mean G = (4* 25)1/2

=100 ½ =10Thus 4, 10 and 25 are in GP

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SummaryProgressionisapatternofnumberswhereinaquantityincreasesordecreasesprogressivelybyfixedamount.•Asuccessionofnumbersformedandarrangedinadefiniteorderaccordingtocertaindefiniterule,iscalleda•progression.If quantities increase or decrease by common difference, then they are said to be in arithmetical progression •(AP).a, a+d, a+2d, a+3d… are the successive terms of an arithmetic progression.•The n• th term of an AP is given by

= a + (n-1) dTheSumoffirstnterms(• ) of an AP is usually denoted by and is given by

= (2a+ (n-1) d)The n• th term Tn of a GP is given by Tn = a(r) n-1

ThesumoffirstntermsofaGPisusuallydenotedbyS• n and is given by Sn =

ReferencesGuy, R. K., 1994. • Unsolved Problems in Number Theory, 2nd ed., Springer-Verlag.Hardy, G. H. and Wright, E. M., 1979. • An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, New York.Prof. Chakraborthy, • Introduction to the theory of Probability, [Video Online] Available at: <http://www.youtube.com/watch?v=r1sLCDA-kNY> [Accessed 31 August 2012].Penfield,P.,2010,• Probability, [Video Online] <http://www.youtube.com/watch?v=_oBgnTy85fM> [Accessed 31 August 2012].Fill, J., • Journal of theoretical Probability, [Online] Available at: <http://www.springer.com/mathematics/probability/journal/10959> [Accessed 31 August 2012].Khan, • Probablity, [Online] Available at: <http://www.khanacademy.org/math/probability/v/basic-probability> [Accessed 31 August 2012].

Recommended ReadingWaters, D., 2006. • Quantitative Methods for business, Progression, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative methods, Arithmetic Progression, Elsevier.Slater, J, C., 2007. Quantitative Methods, Progression, Thomson Learning.•

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Self Assessment________isapatternofnumberswhereinaquantityincreasesordecreasesprogressivelybyfixedamount.1.

Progressiona. Meanb. Geometricc. Arithmeticd.

Asuccessionofnumbersformedandarrangedinadefiniteorderaccordingtocertaindefiniterule,iscalleda2. _______ .

meana. arithmeticb. geometricc. progressiond.

If quantities increase or decrease by common difference, then they are said to be in ________ progression.3. geometrica. arithmeticb. harmonicc. meand.

The following series is in which progression? Series: 4,8,12,16,20,24…4. geometrica. arithmeticb. harmonicc. meand.

The following series is in which progression? Series: 3,-6, 12,-24, 48,-96, 192…5. geometrica. arithmeticb. harmonicc. meand.

The following series is in which progression? Series: 1/10, 3/20, 1/5, 1/46. if AM is 10, 20/3, 5, 4...

geometrica. arithmeticb. harmonicc. meand.

What is the sum of the series 5/2,4,11/2,7,….upto 21 terms?7. 350a. 360b. 357.60c. 367.60d.

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Whatisthesumoftheinfiniteseries1,0.5,0.25…?8. 1a. 2b. 3c. 4d.

What is the 749. th term of the series 3, 7, 11, 15…?265a. 285b. 295c. 300d.

Whatisthesumoffirst6termsoftheseries3,9,27,81….?10. 1092a. 1000b. 1900c. 1192d.

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Chapter V

Probability

Aim

The aim of this chapter is to:

defineprobability•

expalin the concept of sample space and events•

elucidate conditional probability•

Objective

The objective of this chapter is to:

explain rules of probability•

discuss Bayes theorem•

discuss types of events•

Learning outcome

At the end of this chapter, you will be able to

understand the ways to solve probability•

enlist the steps to solve probability•

underst• and the formulae to calculate conditional probability

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5.1 IntroductionProbability is the method to determine the chance of happening of a particular event in a random experiment. Example: probability of a good monsoon.

5.2 Definitions

5.2.1 ExperimentAnoperationwhichcanproducesomewelldefinedoutcomesiscalledanexperiment.

5.2.2 Deterministic ExperimentDeterministicexperimentistheonewhichgivesacertaindefiniteresult.example:acidisaddedtoabase.

5.2.3 Random ExperimentRandom experiment is the one which gives one or more results under identical conditions. An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance is called a random experiment.

5.2.3.1 Examples of Performing a Random Experiment

rolling an unbiased dice•tossing a fair coin•drawingcardfromapackofwellshuffledcards•picking up a ball of certain colour from a bag containing balls of different colours•

5.2.3.2 Details

When we throw a coin, then either head (H) or a tail(T) appears•A dice is a solid cube, having 6 faces ,marked 1,2,3,4,5,6 respectively, when we throw a die, the outcome is the •number that appears on its upper faceA pack of cards has 52 cards; it has 13 cards of each suit namely spades,clubs,hearts and diamonds; cards of •spades and clubs are black cards; cards of heart and diamonds are red cards; there are four honours of each suit, they are Aces,Kings,Queens and Jacks; these are called face cards

5.2.3.3 Sample Space :The set of all possible outcomes of a random experiment is known as the sample space and every outcome is a sample point. When we perform an experiment, then the set S of all possible outcomes is called the sample space.

Examples of Sample SpaceIn tossing a coin,S=H,T•If two coins are tossed ,then S=HH,HT,TH,TT•In rolling a dice , we have S=1,2,3,4,5,6•

5.2.4 Elementary EventAneventisasubsetofthesamplespace.Aneventdefinedbysingletonsetiscalledanelementaryeventorasimpleevent.

5.2.5 Impossible EventTheeventdefinedbyanemptysetiscalledanimpossibleevent.

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5.2.6 EventsEvents are said to be mutually exclusive these events cannot occur simultaneously.

5.2.7 Mutually Exclusive EventTwo or more event are said to be mutually exclusive if these events cannot occur simultaneously.

5.2.8 CompatibilityTwo or more events are said to be compatible if they can occur simultaneously.

5.2.9 Independent EventsTwo or more events are said to be independent if the happening or non-happening of any event does not affect the happening of others. Events are no way related to each other.

5.2.10 Dependent EventsTwo or more events are said to be dependent if the happening or non-happening of any event does affect the happening of others. Events are related to each other.

5.3 ProbabilityIf A is an event of the sample space S, thenProbability of A=P (A) =Number of cases favourable for A /Total number of casesWhere

0• ≤P (A) ≤1P (A) +P (A’) =1•

If ‘a’ cases are favourable to A and ‘b’ case are not favourable to A thenP (A) =a/ (a+b); P (A’) =b/ (a+b)

We say that the odds in favour of A are a: b i.e. a to b and odds against A are b: a i.e. b to a. Thus if an event can happen in ‘a’ ways and fail in ‘b’ ways, and each of these is equally likely, the probability or chance of its happening is a/ (a+b) and that of its failing is b/ (a+b).The chance of happening of an event ‘a’ is also stated as “the odds are a to b in favour of the event, or b to a against the event”.

If ‘p’ is the probability of happening of an event then ‘(1-p)’ is the probability of its not happening.

5.3.1 Probability of Occurrence of an EventLet S be the sample space and let E be an event.Then E ⊆ STherefore, P (E) =

5.3.2 Results on Probability

P(S) =1•0• ≤ P(E)≤ 1P(• φ)= 0For any events A and B ,we have P(A • ∪ B) =P(A)+P(B)-P(A ∩ B)IfĀdenotes(notA),thenP(Ā)=1-P(A)•

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5.3.3 Binomial DistributionIf n trials are performed under the same condition and probability of success in each trial is p and q=1-p, then the probability of exactly r successes in n trials is:P(r) = n

5.3.4 Geometric TheoremIfanexperimentisperformedindefinitelyunderthesameconditionsthentheprobabilityoffirstsuccessinthenthtrial is given byP (n) =PWhere p =probability of success in each experimentq is the probability of not getting success =1-p

if and are the chances that the two independent events will happen separately then the chance that they will both happen is x .Also,thechancethatthefirsteventhappensandthesecondfailsis x (1- ) and so on.

If an event can happen in two ways which are mutually exclusive and if and are the two respective probabilities then the probability then the probability that it will happen in some one of these ways is + .

5.4 Conditional Probability

5.4.1 Conditional probability of Dependent EventsConsider the three events. Two dies are tossed thenA: sum of the scores is evenB: sum of the scores is less than 6WritingthesamplespaceitcanbeseeneasilythattheprobabilityofeventAis½andthatofeventBis5/18.Butifit is given that the event A has already occurred then the probability that sum of the scores is less than 6 is 2/9.

The event B is same but its probability changes accordingly A has occurred or not. Thus the probability of an event B when event A has already occurred is denoted by P (B|A)andisdefinedby

P (B | A) = and P (A | B) = where A ∩ B is A and B.

5.4.2 Conditional probability of Independent EventsWedefinetheconditionalprobabilityofeventA,giventhatBhasoccurred,incaseofAandB,astheprobabilityof event A P (A | B) =P (A)

5.5. Multiplication Rule

5.5.1 Independent EventsThe joint probabilities of two independent events are equal to the product of their individual probabilities.P (A and B)= P(A).P(B)

5.5.2 Dependent EventsA joint probability of two dependent events A and B is equal to probability of A multiplied by probability of B, given that A has occurred.P (A and B) =P (B).P (A | B)

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This formula is derived from the formula of conditional probability of dependent eventsP (A| B) = P (A and B)/P (A)P (A and B) =P (A| B) .P (A)

5.6 Steps to Solve ProbabilityDefineevents•Find the total outcome of the experiment•Find the probability of each event•If the words ,”either” “or” are used check whether the events are mutually exclusive or not to apply addition •ruleIf the words “both” “and” are used check whether the events are independent or dependent to apply proper •multiplication rule

5.7 Bayes TheoremThomas Bayes addressed both the case of discrete probability distributions of data and the more complicated case of continuous probability distributions.

In the discrete case, Bayes theorem relates the conditional and marginal probabilities of events A and B, provided that the probability of B does not equal to zero.

Each term in Bayes theorem has conventional name:P (• A) is the prior probability or marginal probability of A. It is “prior” in the sense that it does not take into account any information about B.P (• A|B) is the conditional probability of A, given B. It is also called the posterior probability because it is derived fromordependsuponthespecifiedvalueofB.P (• B|A) is the conditional probability of B given A. It is also called the likelihood.P (• B) is the prior or marginal probability of B, and acts as a normalising constant.

Bayes’ theorem in this form gives a mathematical representation of how the conditional probability of event A given, B is related to the converse conditional probability of B given A.P (A| B) = P (B|A).P (A)/P (B)

Solved ExamplesExample 1Inathrowofacoin,findtheprobabilityofgettingahead.

Solution:Here S= H, T and E= H

Therefore P (E) = =1/2

Example 2Two unbiased coins are tossed .What is the probability of getting at most one head?

Solution: Here, S = HH, HT, TH, TTLet E= event of getting at most one headE= TT, HT, THTherefore P (E) = =3/4

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Example 3An unbiased die is tossed .Find the probability of getting a multiple of 3.

Solution:Here S= 1, 2, 3, 4, 5, 6Let E be the event of getting a multiple of 3.Then, E= 3, 6

Therefore P (E) = =2/6 =1/3

Example 4Inasimultaneousthrowofapairofdice,findtheprobabilityofgettingatotalmorethan7.Here, n(S) = (6 x 6) =36

Let E= event of getting a total more than 7=(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6) P (E) = = =5/12

Example5A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same colour.

Let S be the sample space .Then,n(S)= Number of ways of drawing 2 balls out (6+4)= = 10 x 9 /2 x 1 =45Let E =event of getting both balls of the same colour. Then,n(E)=Number of ways of drawing (2 balls out of 6) or (2 balls out of 4)

= 6 +4 = + = (15+6) =21

P (E) = =21/45=7/15

Example 6Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6?

Clearly n(S) =6 x 6 =36

Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6 .ThenE=(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2)(4,4)(5,1)(5,3),(6,2),(6,6)n (E)=14

Hence P (E) = =14/36 = 7/18

Example 7Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are queens?

Solution:We have n(S) = 52 = =1326Let A = event of getting both black cards, B=event of getting both queensA ∩ B = event of getting queens of black cards.

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n (A) = 26 = (26 x 25)/ (2 x1) =325n (B) =4 = (4 x 3)/(2 x 1)=6and n (A ∩ B)= 2 =1P (A) =n(A)/n(S) =325/1326P (B) =n(B)/n(S) =6/1326P (A ∩ B) = n (A ∩ B)/n(S) = 1/1326Therefore P (A ∪ B) =P (A) +P (B)-P (A ∩ B)= (325/1326 + 6/1326 -1/1326) =330/1326=55/221

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SummaryProbability is the method to determine the chance of happening of a particular event in a random experiment.•Anoperationwhichcanproducesomewelldefinedoutcomesiscalledanexperiment/event.•Random experiment is the one which gives one or more results under identical conditions. An experiment in •which all possible outcomes are known and the exact output cannot be predicted in advance is called a random experiment. The set of all possible outcomes of a random experiment is known as the sample space and every outcome is •a sample point.Two or more event are said to be mutually exclusive if these events cannot occur simultaneously.•Two or more events are said to be independent if the happening or non-happening of any event does not affect •the happening of others. Events are no way related to each other.P (E) = •P (B • | A) = In the discrete case, Bayes theorem relates the conditional and marginal probabilities of events • A and B, provided that the probability of B does not equal to zero.

ReferencesGrinstead, C. M. and Snell, J. L., 1997. • Introduction of Probability: Probability, AMS Bookstore, pp133-137.Mosteller, F., 1987. • Probability: Probability, 1st ed., Dover Publications.Prof. Chakraborthy, • Introduction to the theory of Probability, [Video Online] Available at: <http://www.youtube.com/watch?v=r1sLCDA-kNY> [Accessed 31 August 2012].Penfield,P.,2010,• Probability, [Video Online] <http://www.youtube.com/watch?v=_oBgnTy85fM> [Accessed 31 August 2012].Fill, J., • Journal of theoretical Probability, [Online] Available at: <http://www.springer.com/mathematics/probability/journal/10959> [Accessed 31 August 2012].Khan, • Probablity, [Online] Available at: <http://www.khanacademy.org/math/probability/v/basic-probability> [Accessed 31 August 2012].

Recommended ReadingWaters, D., 2006. • Quantitative Methods for Business, Probability, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative Methods, Probability, Elsevier.Slater, J, C., 2007. • Quantitative Methods, Probability, Thomson Learning.

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Self Assessment__________ is the method to determine the chance of happening of a particular event in a random 1. experiment.

Probabilitya. Experimentb. Eventc. Deterministicd.

Anoperationwhichcanproducesomewelldefinedoutcomesiscalledan_________.2. probabilitya. experimentb. eventc. deterministicd.

_________ experiment is the one which gives one or more results under identical conditions.3. Deterministica. Randomb. Independentc. Dependentd.

The set of all possible outcomes of a random experiment is known as the _________.4. eventa. sample spaceb. sample pointc. experimentd.

An event is a __________ of the sample space.5. subseta. setb. proper subsetc. probabilityd.

Two or more events are said to be ___________ if the happening or non-happening of any event does affect 6. the happening of others.

independenta. mutually exclusiveb. elementaryc. dependentd.

Two or more event are said to be _________ if these events cannot occur simultaneously.7. independenta. dependentb. mutually exclusivec. elementaryd.

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Two unbiased coins are tossed .What is the probability of getting at most one head?8. 1/2a. 1/4b. 3/4c. 5/2d.

An unbiased die is tossed .What is the probability of getting a multiple of 3?9. 2/3a. 5/6b. 1/3c. 1/6d.

A bag contains 6 white and 4 black balls. Two balls are drawn at random. What is the probability that they are 10. of the same colour?

1/15a. 4/15b. 2/15c. 7/15d.

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Chapter VI

Permutations and Combinations

Aim

The aim of this chapter is to:

introduce the concept of permutation•

definecombination•

explain the principles of counting•

Objective

The objectives of this chapter are to:

explain the fundamental principles of counting•

elucidate the principle of addition•

discuss the principle of multiplication•

Learning outcome

At the end of the chapter, you will be able to:

understand the permutation operations in different scenario•

identify the combination operations in different scenario•

understand factorial notation•

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6.1 IntroductionPermutation is the method that helps us to determine the number of ways a group of objects can be selected and then arranged but combination only tells us the number of ways the objects can be selected.

Example: Tossing a coin, placing a ball in a box, rolling a dice selecting a person from a crowd are all physical processes that have a number of possible outcomes. A ball can be placed in a box in one way, the two possible outcomes 1,2,3,4,5 and 6,selecting a committee of four from several hundred people has many outcomes. When considering such physical processes we use permutations and combination to easily arrive at the correct answer.

6.2 Basic Calculation Used

6.2.1 Factorial Notation

theproductoffirstnnaturalnumbersisdenotedbyn!andisreadas“nfactorial”or“factorialn”.•bydefinition0!=1•wecanrepresentn!as;n!=nx(n-1)!•Thevalueofn!=1x2x3x....xn•Suppose we have n x (n-1) x (n-2) x ....x(n-r+1) = •Where r is a natural number less than or equal to n

IllustrationLet us take “Represent 10 x 9 x 8 in factorial notationWehave10!=10x9x8x7x6x5x4x3x2x1Wehave7!=7x6x5x4x3x2x1

Therefore 10 x 9x 8 = = =

6.3 Fundamental Principles of Counting

6.3.1 Principle of AdditionIf one thing can be done in m different ways and a second thing can be done in n different ways independent of thefirst,theneitherofthemcanbedonein(m+n)differentways.

Example:If in a toy shop there are 10 different toy guns and 15 different toy dolls. Then the number of ways to select or to choose a toy from the shop is 10 +15 =25.Because we have to choose a toy, which could be either a toy gun or a toy doll.

6.3.2 Principle of MultiplicationIf one thing can be done in m different ways and a second thing can be done in n different ways, independent of the first,thenboththethingscanbedonethethingscanbedonetogetherin(mxn)differentways.OrIfonetaskcanbedividedinto2subpartswherethefirstsubpartcanbecompletedinmwaysandsecondsubpartcan be completed in n ways then the total number of different ways to complete the task is m x n.

IllustrationIf a Person wants to go from Mumbai to Nagpur via Pune, there are 6 different ways to go from Mumbai to Pune and 6 different ways to go from Pune to Nagpur we can say person has 6 x 6 different ways.

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Route 1

Mumbai Nagpur

Pune Route A

Route 2 Route B

Route 3 Route C

Route 4 Route D

Route 5 Route E

Route 6 Route F

Fromtheabovefigure,wecanseethatapersoncangofromMumbaitoNagpurbythefollowingroutes:Route 1-Route ARoute 1-Route BRoute 1-Route CRoute 1-Route DRoute 1-Route ERoute 1-Route F

Similarly by choosing route 2,3,4,5 or 6 different ways for each of the routes thus the total number of different ways a person can go from Mumbai to Nagpur is 6 x 6 =36 different ways.

Generalization of Fundamental Principle of Counting If n different things can be done in m1,m2,m3,...,mn different ways respectively, independent of each other then

Any one of them can be done in ma. 1+m2+m3+...+mn different waysAll of them can be done in the same order in mb. 1 x m2 x....x mn different ways.

6.4 PermutationPermutation is the number of arrangements of given objects.

Illustration:If Raj, Ram and Shiv are three people to be seated for a Photograph. The number of ways they can be seated are Raj, Ram, ShivRaj, Shiv, RamRam, Raj, ShivRam, Shiv, RajShiv, Raj, RamShiv, Ram, Raj

Therefore,thetotalnumberofarrangementsis6or3!Soiftherearenpeopleorobjectswhichwehavetoarrangethenthetotalnumberofarrangementsisn!

Permutations of n distinct objects when only r are taken at a time and arranged in a straight line, then the number of arrangements is given by n P r or P(n,r),where

n P r =

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6.4.1 Basic Forms of PermutationsFollowing are the basic forms of permuataions6.4.1.1 All given Objects are DistinctNumber of Permutations when all the given n distinct objects(r=n) are taken

n P n = = =n![As0!=1]

Illustration: All given objects are distinct7 people are there in a meeting In how many ways they can sit in a row of 7 chairsHereall7peoplehavetobeseatedthereforenumberofarrangement=7!=5040

6.4.1.2 When k cannot be SelectedPermutations of n distinct objects when only r objects are taken at a time and arranged in a straight line such that k of the given objects is never taken.

Therefore the number of objects left are (n-k) then the the number of arrangements is given by (n-k) P r =

Illustration: When K cannot be selectedThere are 3 varieties of mangoes(A,B,C),4 different varieties of Apples(D,E,F,G) and 7 different varieties of grapes(H,I,J,L,M,N).In how many ways we can arrange 5 fruits on a table when G,M,N should not be taken.

In total we have 3 +4+7=14 fruits. (n=14 k=3 because 1 variety of apple, 2 varieties of grapes cannot be taken and r=5 as we have to arrange 5 fruits.

(n-k) P r = = = = =55440

6.4.1.3 When all the given n objects are not distinctWhen all the given n objects are not distinct. Out of those n objects p of them are of one kind ,q of them are of second kind and r of them of third kind and the remaining are distinct objects then number of permutations are given by

Illustration: When all of the objects are not distinctWhen all of the objects are not distinctThere are 3 mangoes, 4 apples and 7 other fruits one of each. In how many ways we can arrange fruits on a table.In total, we have 3+4+7 =14 fruits (n=14, P=3 mangoes q=4 apples The number of arrangements is =

6.4.1.4 Circular Permutation Numberof permutationsof ndistinct objects taken all at a time in a circle is (n-1)!.If the arrangement is notdependent upon direction i.e. either clockwise or anti-clockwise the arrangements will remain same then the number of permutations =

Illustration: Circular Permutation There are 3 people A, B, C.The number of ways they can be seated around a round table. Therefore number of ways =(n-1)!=(3-1)!=2!=2

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If instead of three people we have 3 beads to make a necklace then the number of arrangements become = = =1 because if we reverse the necklace then the order 1 is same as 2.Therefore only one arrangement is possible.

6.4.1.5 Repetition is AllowedIf repetition is allowed then the number of permutations for n distinct objects from which r are taken at a time nr.

Illustration: Repetition is allowedIn a particular country the pin code for different cities are given by using digits from 1 to 8 and the size of pin code is 4 digits. Then how many different pin codes are possible.n=8 and r=4 hence nr=84=4096.

6.5 CombinationCombination is the number of selections out of given objects.

Illustration: CombinationIf Manish, Harish and Jaideep are three people out of which 2 have to be selected for a photograh.The number of ways they can be selected areManish, HarishHarish, JaideepManish, Jaideep

The total number of selections is 3 or 3C2.

Combination of n distinct objects when only r objects are taken at a time, then the number of selections is given by nCr or C(n,r) where n is the number r is the number of object selected.

General Formula is nCr =

6.6 Basic Forms of Combination

6.6.1All Given Objects are distinct Number of Combination when all the given n distinct objects (r=n) are taken, then

n C n = = =1

Illustration:All given objects are distinct7 people are there in a company.In how many ways they can be a part of a meeting of 7 People?Here all 7 people are selected .Therefore number of selections =1

6.6.2 When K objects cannot be selectedCombination of n distinct Objects when only r are taken at a time such that k of the given objects is never taken. Therefore number of objects left are n-k then the number of selections is given by n-k C r =

Illustration: When K objects cannot be selected There are 3 varieties of mangoes(A,B,C),4 different varieties of apples(D,E,F,G) and 7 different varieties of grapes(H,I,J,K,L,M,N).In how many ways can 5 fruits be selected when we cannot take G,M,N.

In total we have 3+4+7 =14 fruits. Therefore n=14, k=3 because 1 variety of apple, 2 varieties of grapes cannot be selected and r=5 as we have to select 5 fruits.

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n-k C r = = = = = 462

6.6.3 When k Objects are always Selected Combination of n distinct objects when r objects are taken at a time such that k objects are always selected. Then the total number of selections = n-k C r-k

Illustration: When k objects are always selectedThere are 11 players in a hockey team. In how many ways we can select 5 players for the board meeting if captain and vice captain have to be there.Here n=11, k=2 (captain and vice captain),r=5Total number of way the members for the meeting can be selected = n-k C r-k= 11-2 C 5-2 =9 C 3 = =84

6.6.4 Distribution of Objects into two GroupsIf there are (m+n) objects and we required to divide them into groups of m and n objects respectively. Then the number of ways will be m+n C m or m+n C n

Illustration: Distribution of Objects into two groupsThere are 10 people in a company. The company wants to divide them into two teams such that one team has 6 people and the other team has 4 people. With the condition that same person cannot be a part of both the teams and each person has to be a part of either of the teams.

Number of different ways to select the two teams = m+n C m = 6+4 C 6 = =210

Note:When m= n then the number of ways =

6.6.5 Distribution of Similar ObjectsIf we have n same objects and want to distribute them amongst r persons then the total number of ways to distribute them is given by n+r-1 C r-1

Illustration: Distribution of similar objectsIf we have 10 chocolates with us and want to distribute them amongst 4 students then the number of different ways to distribute them is= n+r-1 C r-1 where n=10 and r=4Therefore, n+r-1 C 4-1 =13 C 3=286

Note:If each of the people is getting atleast one then the number of ways to distribute will be n-1 C r-1

6.6.6 Total possible Combination of n Distinct Objects If we have n distinct objects, then the total number of ways to select an object or a group of objects from them is given by 2n-1

Illustration: Total possible combination of n distinct objectsIf a person has seven friends, in how many ways can he invite one or more of them to attend a tea party?

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He can invite one or two or seven friends in

7C1+7C2+7C3+....+7C7 =27-1=127 ways

6.6.7 When All are not Distinct ObjectsIf there are p+q+r objects where p are of one type are of second type and r are distinct objects then the number of ways to select atleast 1 object out of them is (p+1)(q+1)2r-1

Illustration: When all are not Distinct ObjectsIn a library there are 4 science books, 8 mathematics books and 10 books of other different subjects then the number of ways in which a person can select atleast 1 book from any subject is given by (p+1)(q+1)(2r)-1=(4+1) (8+1)210 -1 = 5 * 9 * 1024 -1 =46079

6.6.8 When all are Distinct but of Different KindIf there are p+q+r objects where p are distinct objects of one type ,q are distinct objects of second type and r are distinct objects of third type then the number of ways to select atleast 1 object of each type out of them is (2p-1)(2q-1)(2r-1)

Illustration: When all are distinct but of different kindSuppose if a person has 5 blue shirts of different shades, 3 black trousers of different shades and 4 grey ties of differentshadesnowinhowmanyhecanwearanoutfitforaparty.

Thereforep=5,q=3andr=4.Thenumberofdifferentwayshecanselectanoutfitfortheparty=(25-1)(23-1)(24-1) =31 * 7 *15 = 3255.

6.7 Special Case(Permutation and Combination Simultaneously)If there are n objects we have to arrange r out of those n such that k objects are always selected for the arrangement. Then total number of arrangements are given by n-k C r-k*r!

Illustration:There are 10 different fruits and we want to arrange 4 fruits on the table such that 2 of them are always on the table.

We know that n=10 fruits r=4 k=2Thetotalnumberofarrangementsaregivenbyn-kCr-k*r!=8C2*4!=672.

6.8 Basic Manipulation on Permutation and Combinationsn C • n-r =n C r

If n C • p = n C q then either p=q or p+q =nn P • n= n P n – 1

n P • r = n C r*r!n P • r = n-1 P r + r n-1 P r-1

n C • r + n C r-1 = n+1 C r

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Solved Examples

Example 1:Fromapackof52cardsfindthenumberofwaysinwhich

a king or a queen can be drawna. both a king and a queen can be drawnb.

Solution & Explanation:A king can be drawn in 4 different ways; a queen can be drawn in 4 different ways. By fundamental principle of multiplication a king and a queen can be drawn in 4 x 4 =16 ways.

Example 2:How many words can be formed with the letters of the word Mathematics? In how many of them do the vowels occur together?

Solution & Explanation:Total letters =11, M, A, T occur twiceTotalarrangements=11!/(2!2!2!)=4989600

Treat the four vowels A, A, E, I as one unit. This with remaining 7 letters of which two are repeated can be arranged in8!/2!2!Ways.Fourvowelscanbearrangedamongthemselvesin4!/2!Ways.ThereforeTotalnumberofways=(8!/2!2!)(4!/2!)=120960

Example 3:Twenty persons were invited for a party. In how many ways can they and the host be seated at a circular table? In how many of these ways will two particular persons be seated on either side of the host?

Solution & Explanation:There are 20 +1 persons to be seated at the table .Fixing the seat of one person the remaining 20 can be seated in 20!ways.

Twoparticularpersonscanbesittingoneithersideofthehostin2!Ways.Remaining18canarrangethemselvesin18!Ways.

Thereforetherequirednumber=2!*18!

Example 4:Tendifferentlettersaregiven.Wordswithfivelettersaretobeformedfromthesegivenletters.Findthenumberofwords which have at least one letter repeated.

Solution & Explanation:When there is no repetition of letters,Total number of words with 5 letters = 10 P 5 =30240 When any letter is repeated any number of times, total number of words =10 5

Therefore required number of words =100000-30240=69760

Example 5A photograph of 4 players is to be taken from 11 players of a cricket team. How many different photographs can be taken if in each photograph captain and vice captain

must be included a. are never includedb.

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Solution & Explanation:Two players can be chosen in 9 C 2 ways

Total number of photographs = 9 C a. 2*4!=864Total number of photographs =9 P b. 4 ways

Example 6From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking atleast one green and one blue ball?

Solution & Explanation:Atleastonegreenballcanbechosenfromfivegreenballsin25-1=31 waysAt least one blue ball can be chosen from four blue balls in 24-1 =15 waysAt least one or no red ball can be chosen in 23=8 ways

Therefore by generalization of fundamental principle, required number of ways =31 * 15 *8 =3720

Example 7From 3 pears, 4 oranges and 5 apples, how many selections of fruits can be made by taking

at least one of themi. at least one of each kindii.

Solution & Explanation Out of 3 pears, we may take either 0 or 1 or 2 or 3 .Thus pears can be chosen in 4 different ways .Similarly, iii. oranges in 5 and apples in 6 ways. Number of ways of selection of fruits =4 *5*6 =120.But this also includes the case when no fruit is taken. Rejecting this case, the required number of ways=120-1=119. At least one pear is to be chosen, we may take either 1, 2 or 3 pears. Thus pears can be chosen in 3 different iv. ways. Oranges can be chosen in 5 different ways. Apple can be chosen in 5 different ways

Therefore total number of ways = 3 * 4 * 5=60

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SummaryPermutation is the method that helps us to determine the number of ways a group of objects can be selected and •then arranged but combination only tells us the number of ways the objects can be selected.n P • r = Permutations of n distinct objects when only r objects are taken at a time and arranged in a straight line such •that k of the given objects is never taken.(n-k) P • r = When all the given n objects are not distinct. Out of those n objects p of them are of one kind ,q of them are •of second kind and r of them of third kind and the remaining are distinct objects then number of permutations are given by Numberofpermutationsofndistinctobjectstakenallatatimeinacircleis(n-1)!•Combination is the number of selections out of given objects. General Formula is nC• r = Combination of n distinct Objects when only r are taken at a time such that k of the given objects is never taken. •Therefore number of objects left are n-k then the number of selections is given by n-k C r = If we have n distinct objects, then the total number of ways to select an object or a group of objects from them •is given by 2n-1.

ReferencesKumar, K. R., 2005. • Discrete Mathematics, Permutation and Combination, Firewall Media, p23-47.Rao, G. S., 2002• . Discrete Mathematical Structure, Permutation, New Age International, pp14-54.Nix, R., • Probability, Combinations and Permutations, [Online] Available at: < http://www.chem.qmul.ac.uk/software/download/qmc/ch5.pdf> [Accessed 31 August 2012].Permutations and combinations• , [Online] Available at: < http://download.nos.org/srsec311new/L.No.07.pdf> [Accessed 31 August 2012]Kirthivasan, K., • Permutation and combination, [Video Online] Available at: < http://www.youtube.com/watch?v=Dsi7x-A89Mw> [Accessed 31 August 2012].IIT JEE, • Permutation and combination, [Video Online] Available at: < http://www.youtube.com/watch?v=b8mvJm1vvdQ> [Accessed 31 August 2012].

Recommended ReadingWaters, D., 2006. • Quantitative Methods for Business, Permutation and Combination, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative Methods, Permutation, Elsevier.Slater, J, C., 2007. • Quantitative Methods, Combination, Thomson Learning.

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Self Assessment________ is the number of arrangements of given objects.1.

Permutationa. Combinationb. Factorialc. Selectiond.

__________ is the number of selections out of given objects.2. Permutationa. Combinationb. Factorialc. Selectiond.

__________ is the method that helps us to determine the number of ways a group of objects can be selected 3. and then arranged.

Permutationa. Combinationb. Factorialc. Selectiond.

__________ only tells us the number of ways the objects can be selected.4. Permutationa. Combinationb. Factorialc. Selectiond.

There are 3 varieties of mangoes(A,B,C),4 different varieties of Apples(D,E,F,G) and 7 different varieties of 5. grapes(H,I,J,L,M,N).In how many ways we can arrange 5 fruits on a table when G,M,N should not be taken?

22330a. 33440b. 44550c. 55440d.

In how many ways they can be a part of a meeting of 7 People?6. 5040a. 3040b. 2040c. 1040d.

There are 11 players in a hockey team. In how many ways we can select 5 players for the board meeting if 7. captain and vice captain have to be there.

81a. 80b. 83c. 84d.

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7 people are there in a company. In how many ways they can be a part of a meeting of 7 People?8. 7!a. 7b. 5040c. 1d.

From 5 different green balls, four different blue balls and three different red balls, how many combinations of 9. balls can be chosen taking atleast one green and one blue ball?

3120a. 3720b. 3520c. 3420d.

From a pack of 52 cards in how many number of ways in which a king or a queen can be drawn?10. 2a. 1b. 4c. 3d.

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Chapter VII

Interpolation

Aim

The aim of this chapter is to:

defineinterpolation•

introduce the applications of interpolation•

describe the importance of interpolation•

Objectives

The objectives of this chapter are to:

explain the different methods of interpolation•

elucidate Newton-Gauss forward and backward method•

explicate the graphical method of interpolation•

Learning outcome

At the end of this chapter, you will be able to:

understand the need for interpolation•

identify the Newton’s method of advancing differences•

comprehend Lagrange’s method• for calculation interpolation

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7.1 IntroductionInterpolation is the method of statistical estimation and the word literally means making insertions. Simply interpolation is understood by following example;

If we need to know the population of our country, for any intermediary year, says 1985, one logical approach would betoworkforwardfromthepopulationof1981,byaddingbirthsandinflowofthepeopleintothecountryanddeductingdeathsandoutflowofpeoplefromthecountryduring1981-1985.

Thusthedataonpopulationoftheyear1985isrequired,100%accuratefiguresarereallynotrequired.

7.2 Definition of Interpolation“Interpolation”maybedefinedasthetechniqueofobtainingthemostlikelyestimatesofcertainquantityunder•certain assumptions. - D.N. ElhanceInterpolationisastatisticaldeviceusedtoestimatethemostlikelyfigureundercertainassumptionswithinthe•given limitsInterpolation provides us the missing quantity of a series so that we can establish the while extrapolation are •the techniques of obtaining the most likely estimates of certain quantity under certain assumptionsInterpolatedfiguresarenotperfectsubstitutesoftheoriginalfigures.Theyareonlybestpossiblesubstitutes•on certain hypothesis

7.3 ApplicationInterpolationiswidelyusedbybusinessmen,administrators,sociologists,economistandfinancialanalysis.•It helps in completing the incomplete, lost or destroyed records.•Eg. Infinancialanalysis the interpolationused tofindout the IRR(internal rate of return) of a project, all•investment decisions which require to use of the Present value and future value interest factor tables.

7.4 Need and Importance of InterpolationInadequacy of data

Sometimes it is not possible to collect the whole data about the problem under study. Even if it were possible •to collect the whole data it may not be worthwhile to do so due to a large amount of expenditure involved or due toorganisationaldifficultiesTechnique of interpolation can be used for making best estimates, at the least cost•Theseestimateswillbeamoreusefulfigurethanroughestimates•

To estimate intermediate valueIn certain cases data is collected after long intervals•For example; in India the census of population is collected after every ten yrs•The techniqueof interpolationwillbeneeded toestimate thefigureofpopulation for intermediate• years

Lost dataSometimesthedataislostduetofire,earthquakeetc.Theinterpolationtechniqueishelptofillthegaps• in statistical information due to lost data

Uniformity ofsataStatistics concerning a particular phenomenon are collected by different agencies, it destroys its •uniformity.In such cases comparison of data becomes difficult.So to establish uniformity of data, the techniquesof• interpolation is used.

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ForecastingThe forecasting activity regarding data is practical utility for economic planning, policy formulation, production •decisions etc

7.5 Methods of InterpolationDifferent methods of interpolation are explained below.

7.5.1 Graphical Method

It is simplest method of Interpolation.•In this method the data is represented in graph, i.e. on X-axis all independent variables and on Y-axis all •dependent variables are taken.The curve is formed after joining the points, this curve give interrelation between two variables.•From the point of X-axis, for which the value of y is to be interpolated, a line parallel to Y-axis will be drawn.•From the point where this line will cut the curve, a line parallel to X-axis will be drawn, here the value of y will •be found from the point where the line cuts Y-axis, this is called Interpolated figure or Value

7.5.2 Newton’s method of advancing differences

This method is applicable for the following cases•The independent variable advances by equal intervals•The value to be interpolated is different from the equidistant value•The value to be interpolated lies in the beginning of the data•Thismethodisknownasfiniteoradvancingdifferencesmethodbecauseafterfindingoutdifferencesinthe•values of y, the process is extended further till only one difference remainsWe also taken into consideration that the +ve and -ve sign while calculating the differences•

7.5.3 Lagrange’s Method

Lagrange’s interpolating polynomial is another very good formula for interpolation•This method has no restriction on the x-variable whether it should be equally spaced or not•This method can be used for any value of x either for interpolation•It is also to • estimate the argument of x for given value of y, it means the Lagrange;s formula can be used for inverse interpolation also.The only demerit of Lagrange’s formula is that, it required heavy computational work

7.5.4 Newton-Gauss Forward MethodIt is method which is used in particular situation. It is used when the independent variable (a) advance by equal intervals (b) the value to be interpolated falls in the middle of the series.

The formula is as under;Gauss Forward Formula is fp=f0+pδ1/2 +G2 δ0

2+G3δ1/2 3 +....

for p ∈ [0,1],where δ is the central difference and G2n =

G2n+1 = where isthebinomialcoefficient

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7.5.5 Newton-Gauss Backward Method

This is also known as Newton Gregory backward formula•Its applicability is described below•When the argument x advances with the equal jumps•When the x-value to be interpolated lies near the end of the series•Diagonal difference table is used in Newton’s backward formula, but the differences are used in reverse •order.

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SummaryAlgebraic expressions in which the variables concerned have only non-negative integral exponents are called •polynomialsThe standard form of a polynomial in one variable is that in which the terms of the polynomial are written in •the decreasing order of the exponents of the variableInterpolation provides us the missing quantity of a series so that we can establish the while extrapolation are •the techniques of obtaining the most likely estimates of certain quantity under certain assumptionsMethods of Interpolation are Graphical method, Lagrange’s Method, Newton-Gauss Forward Method, Newton’s-•Gauss Backward method. etc.Lagrange’s interpolating polynomial is another very good formula for interpolation•This method has no restriction on the x-variable whether it should be equally spaced or not•Gauss Forward Formula is •fp=f0+pδ1/2 +G2 δ0

2+G3δ1/2 3 +....

ReferencesJain, T. R.. and Sandhu, A. S., 2006-07, • Quantitative Methods: Interpolation, VK Publication, pp 7.1-7.38.Agarwal, B. R., 2007. • Programmed Statistics, Interpolation, 2nd ed., New Age International, pp 405-425.Kumar, S., 2008, • Polynomial Interpolation, [Video Online] Available at: < http://www.youtube.com/watch?v=Oy3uudRXolE > [Accessed 31 August 2012].Kumar, S., • Numerical Methods and Programming, [Video Online] Available at: < http://www.youtube.com/watch?v=ZG_TgdyDrf0 > [Accessed 31 August 2012].Interpolation• , [Online] Available at: < http://www.mathworks.com/moler/interp.pdf > [Accessed 31 August 2012].Bourke, P., • Interpolation methods, [Online] Available at: < http://paulbourke.net/miscellaneous/interpolation/ > [Accessed 31 August 2012].

Recommended ReadingWaters, D., 2006. • Quantitative Methods for Business, Interpolation, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative methods, Interpolation, Elsevier.Slater, J, C., 2007. • Quantitative Methods, Interpolation of Polynomials, Thomson Learning.

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Self AssessmentGraphical Interpolation method is_____.1.

Simplea. Algebricalb. Fully reliable c. Reliabled.

Interpolation is helpful in estimating:2. A seriesa. An intermediary value of given argument b. Entry of alternative valuesc. A series of valued .

Lagrange Formula is useful for _____3. Interpolationa. Arithmetic functionsb. Inverse rangec. Inverse extrapolationd.

Lagrange’s polynomials interpolation can be used even if;4. The given arguments are not equally spaceda. Extrapolation is to be doneb. Inverse interpolation is to be donec. Relation to be mappedd.

Interpolation formulae are based on the fundamental assumptions that the data can be expressed as;5. A linear functiona. A quadratic functionb . A polynomial function c. A binomial functiond.

The problems of interpolation are simpler than prediction because;6. Interpolation has fewer restrictions than predictiona. Interpolation is based on more stringent restriction than prediction b. There are no restriction than interpolationc. Itiseasiertofindoutd.

________ is simplest method of Interpolation.7. Graphical methoda. Langrange’s methodb. Interpolation methodc. Inverse methodd.

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In which method the data is represented in graph?8. Langrange’s methoda. Interpolation methodb. Inverse methodc. Graphical methodd.

The value to be interpolated is different from the ______ value.9. equivalenta. equalb. equidistantc. equillibriumd.

_________ provides us the missing quantity of a series.10. Interpolationa. Graphical methodb. Lagrange’s methodc. Newton’s methodd.

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Chapter VIII

Consumer Arithmetic

Aim

The aim of this chapter is to:

defineprofitandloss•

explicate interest•

explain recurring deposit•

Objectives

The objectives of this chapter are to :

discuss types of interest•

elucidate present worth•

describe the concept of discount•

Learning outcome

At the end of this chapter, you will be able

understandimporttermsinprofitandloss•

compare simple and compound interest•

underst• and discount

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8.1 Introduction: Profit and LossIn olden days every one produced or made what they needed. In case there was an excess produced I was an excess produced it was exchanged with others for mutually required goods. The exchange was need based. The system was known as ‘barter’. However, as time passed people found it was better to take produce/goods from a person who was good at producing them.Thus, the system of exchange of goods and services for money began. The introduction ofthemonetarysystemgaverisetotheconceptofgainorlossintheexchangeandthusconceptofprofitandlosswas born.

ExampleA pair of shoes has been bought by the shopkeeper from a cobbler. The cobbler had sold the shoes to the shopkeeper for Rs.150.To make the shoes the cobbler had spent Rs.100 on raw material. The shopkeeper spent Rs.20 on transporting the pair of shoes from the cobbler to the shop. The shopkeeper had also bought a box of Rs 5 as packaging for the pairofshoes.TheshopkeeperfixesthepriceforthoseshoesasRs.300andacustomerpurchasesitafterbargainingfor Rs.250.

The price at which the goods are sold is known as selling price (SP). In the above example Rs.150 is the selling •price for the cobbler and Rs.250 is the selling price for the shopkeeper.The price at which the goods are bought is known as cost price(CP). In the above example the cost price for •the shopkeeper is 175(cost paid to the cobbler+transportation+packaging), while Rs.250 is the cost price for the lady.The cost occurred in relation to the goods apart from the actual cost of the goods is known as overhead charges. •In the above example the overhead charges is Rs.25.(Cost of transportation Rs.20+cost of Packaging Rs.5)Ifsellingpriceismorethanthecostprice,thenitissaidthesellerhasmadeaprofit/gain.Intheaboveexample•theprofitforthecobblerisRs.50(sellingpriceofthecobbler–rawmaterialcostforthecobbler).Theprofitforthe shopkeeper is Rs.75 (selling price of Rs.250-cost price of Rs 175).If the selling price is less than the cost price ,then it is said the seller has made a loss. In the above example if •the shopkeeper had sold the shoes for Rs150 ha would have made a loss(cost price of Rs 175 –Selling Price of Rs.150)The price at which the goods are intended to be sold is known as marked price. In the above example Rs.300 is •the marked price of the shoes .Marked price is also known as the List price or quoted price.When the Selling price is lesser than the marked price, it is said that the seller has given a discount.In the above •example the customer has given a discount of Rs.50 (marked price of Rs.300-selling price of Rs 250).This discount is also known as trade discount.The price at which the goods are sold after discount is known as net price. In the above example the net price •is Rs.250.

8.1.1 Formulae

Profit=sellingprice–costprice•Loss = cost price –selling price•Gain%=(Gain/cost price)* 100•Loss%=(Loss/cost price)*100•If the cost price and gain is given then selling price = [(100+gain%)*cost price]/100•If the cost price and Loss% is given the selling price =[(100-loss%)*cost price]/100•If the selling price and gain% is given that cost price= (100*selling price)/ (100+gain %)•If the selling price and loss% is given then cost price = (100*selling price)/ (100-loss %)•

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The points to remember are as follows:Ifthemarkedpriceisdifferentfromthesellingprice,thentheprofit/lossiscalculatedonthesellingprice.•Percentage gain or loss is calculated on the cost price and not the selling price.•ProfitorlossonRs100(costprice)iscalledprofitorlossPercentagerespectively.•Profitisalsoreferredtoasgain•Discount is always given on the market price.•

8.2 InterestInterest is a fees paid by a borrower of assets to the owner of the assets as form of compensation for the use of assets. It is most commonly the price paid for the use of money or money earned by the deposited funds.

Illustration:When we give our house for someone to live in we get a rent and we get the house back after completion of the lease tenure, similarly when we lend our money to someone we get interest and after completion of the tenure we get our money back.

8.2.1 Terms Used

When a person borrows money he has to pay some extra amount to the lender for using his money. The extra •money paid to the lender is known as Interest.The money that is initially borrowed/lent/invested is known as Principle.•The total sum of principle and interest is known as Amount.•The percentage gain in principle is known as Rate of interest.•The tenure for which the principle is borrowed/lent/invested is known as Time period.•

8.2.2 Simple InterestSimple interest is where the interest is paid to the investor/lender as and when it is due. Only principle is reinvested/renewed.

8.2.2.1 FormulaeLet P=PrincipleA=AmountR=Rate of InterestT=Time periodI=InterestS.I=Simple Interest

ThenS.I = (P*R*T)/100•P= (100*S.I)/(R*T)•R= (100*S.I)/ (P*T)•T= (100*S.I)/ (P*R)•

The points to remember are as follows:For simple interest time period is always calculated in years. If the time period is given in months then it has •to be converted into years.To convert time period from months to year divide the time period by 12 to get the time period in years. Example •if the amount has been given for 5 months then the time period is 5/12 years.Rateofinterestunlessspecifiedistakenasoneyear.•

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8.2.3 Recurring DepositRecurringdepositisadepositinwhichfixedsumofmoneyisdepositedataregularintervalforaspecifiedperiod.Afterthefixedperiodisover,thedepositorgetshisfullmoneybackandaninterest.

The points to remember are as follows:Arecurringschemeislikeapiggybankwhereyoudepositsmallsumsofmoneyatfixedintervalstogeta•consolidated amount at the end of the scheme.Inarecurringschemetheinterestiscalculatedperinterval(monthly,halfyearly,annuallyetc..)atafixedrate.•In a recurring deposit ,interest and the total amount accumulated is given at the end of the scheme period.•In the recurring deposit the amount deposited for the 1• stmonthwillgetaninterestforthefixedperiodsaynmonths.The amount deposited for the 2nd month will get interest for (n-1) months and so on.

8.2.3.1 Formulae

Total principle at maturity=Recurring amount x n(n+1)/2 where n is the period for which the 1• st installment is depositedSimple Interest =•The time period is calculation is taken as 1 month as we have already normalized the principle and hence •t=1/12Total amount at the end of the scheme =Total principle at maturity + interest•

8.2.4 Compound InterestCompound Interest is where the interest is not paid to the investor/lender as and when it is due but is added to the principle at the end of the time period agreed upon. The interest is added to the principle and that becomes the principal for the next tenure.

8.2.4.1 FormulaeLet P =PrincipleA=AmountR=Rate of InterestT=Time periodN=Number of times the interest is due in a yearI=InterestC.I=Compound Interest

Then Amount= P[1+R/100]• NT

C.I =Amount –Principle•Present Worth=A/[1+R/100]• T

Present worth is nothing but the principle invested•Condition 1: Principal = Rs p, time =t years and rate =r% and number of times the interest is due in 1 year =1. •Then the amount after t years is = p[1+r/100]1*t

Condition 2: Principle =rs p; time =t years and a fraction of year[Ex 5 2/5 years] and number of times the interest •is due in 1 year =1.Then the amount after 5 2/5 years = [p(1+r/100)1*5](1+2r/5*100)For the fraction of the year the rate will be rate per annum * fraction of year•Condition 3:Principle =Rs p,Time=t years and rate =r% p.a and number of times the interest is due in 1 year •=3.Then amount after t years is=p [1+r/ (3x100)]3*t

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Similarly, principal=Rs.p, Time =t years and rate =r% p.a and number of times the interest is due in 1 year =2Then amount after t years is =p[1+r/(2 x 100)]2 * t

Condition 4: Principle = Rs p, Time = t years and rate =r• 1%forthefirstyear,r2% for the second year……….rt% for the tth year. The amount after t years is = p(1+r1/100)(1+r2/100)(1+r3/100)……(1+rt/100)

The points to be remembered are as follows:For compound interest time period is always calculated in y years.If the time period is given in months then it •has to be converted into years.Rateunlessspecifiedisalwaysforayear•If the interest is computed more than once in a year then,rate r =rate per annum ( R ) /number of times interest •is computed in a year ( N ).If a sum P is borrowed at r% per annum and compound interest is calculated quarterly or half yearly,then the •compound interest for 1 year is called the effective annual rate.The amount at the end of a certain time period increases as the number of times the compound interest is •calculated in a year increases i.e if two people put the same principle for the same time period on compound interest but ,one computed interest half yearly and the other yearly then the person who is getting interest half yearly and the other yearly then the person who is getting interest half yearly will get more amount at the end of the tenure.

Solved Examples

Example 1Amansellsapenataprofitof20%.Hadheboughtitat20%lessandsolditforRs5less,hewouldhavegained25% .What is the CP of the pen?

Solution:Let the CP be Rs 100; given Gain=20% therefore SP =120New CP=20% less=Rs 80If gain is 25%, then SP= Rs (125*80)/100=Rs 100

Therefore if the difference in SP =Rs (120-100)=Rs 20If the difference in SP is Rs 20, CP is Rs 100

If the difference in SP is Rs 5, CP is Rs (100*5/20) =Rs 25

Example 2IftheSPof10articlesisthesameastheCPof11articles,findthegainpercent.

SolutionLet the CP of 1 article be Re.1Therefore CP of 10 articles=Rs 10SP of 10 articles= CP of 11 articles = Rs 11Therefore gain%=(11-10) x 100/10 =10%

Example 3A man loses 10% by selling a book for Rs 144.What should be his selling price to gain 15?

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SolutionSP=Rs 144 loss=10%Therefore CP=Rs(100 *144)/(100-10)=Rs.160Gain expected = 15%Therefore SP=Rs(115*160)/100=Rs184

Example 4A man sells two radios for Rs 924 each. On one he gains 12% and on the other he loses 12%.How much does he gain or lose on the whole?

Solution:SP of one radio =Rs 924 Gain=12%CP of this radio=Rs (100*924)/112=Rs 825SP of other radio = Rs 924 Loss=12%CP of this radio = Rs (100*924)/88=Rs 1050

Total CP=Rs (825+1050) =Rs 1875,Total SP= Rs (924+924) =Rs 1848

Therefore Loss% = (27 *100)/1875=1.44%

Example 5What will be the simple interest on Rs 625 at 6.5% per annum for 2.5 years?

SolutionS.I=Rs (625 *6.5 * 2.5)/100=Rs.101.56

Example 6A man borrowed Rs 2500 from two money lenders.He paid interest at the rate of 12% per annum for one loan and at the rate of 14% per annum for other .The total interest he paid for the entire year was Rs.326.How much did he borrow at each rate?

SolutionLet he borrow Rs X at 12% and Rs (2500-x) at 14% p.aS.I on Rs x= Rs (x)(12)(1/100) = Rs 3x/25S.I on Rs(2500-x)=Rs (2500-x)(14)(1/100)=Rs (17500-7x)/50 therefore x=Rs 1200 Thus ,he borrows Rs 1200 at 12% and 1300 at 14%.

Example 7Find the compound interest on Rs 4000 for 9 months at 6% per annum,the interest being reckoned i)Quarterly ii)Half yearly

Solution Principle =Rs 4000, Time =9 months=3 quarters rate=6% per annum =6/4=3/2 % per quarter. Therefore i. Amount=Rs[4000 x 1+(3/2)/1003]=4182.71

Compound interest = Rs (4182.71-4000) =182.71Principle=Rs 4000 Time = 9 months =1.5 half years rate= 6% per annum =3% half yearlyii.

Therefore amount Rs[4000(1+3/100)1+3/2)/100]=Rs 4181.80Compound interest =Rs(4181.80-4000=Rs181.80

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SummaryThe price at which the goods are sold is known as selling price (SP).•The price at which the goods are bought is known as cost price(CP).•Ifsellingpriceismorethanthecostprice,thenitissaidthesellerhasmadeaprofit/gain.•If the selling price is less than the cost price ,then it is said the seller has made a loss.•The price at which the goods are intended to be sold is known as marked price.•When the Selling price is lesser than the marked price ,it is said that the seller has given a discount.•Profit=sellingprice–costprice•Loss = cost price –selling price•Gain%=(Gain/cost price)* 100•Loss%=(Loss/cost price)*100•Interest is a fees paid by a borrower of assets to the owner of the assets as form of compensation for the use of •assets.Simple interest is where the interest is paid to the investor/lender as and when it is due. Only principle is •reinvested/renewed.S.I = (P*R*T)/100•Recurringdepositisadepositinwhichfixedsumofmoneyisdepositedataregularintervalforaspecified•period.Compound Interest is where the interest is not paid to the investor/lender as and when it is due but is added to •the principle at the end of the time period agreed upon.Amount= P[1+R/100]• NT

C.I =Amount –Principle•

ReferencesVeena, G. R., 2006. • Business Mathematics, Commercial Arithmetic, New Age International Publishers.Aggarwal, R. S., 2008. • Quantitative Methods, S.Chand Publications.Sadler, W., • Commerce Arithmetic, [Online] Available at: < http://books.google.co.in/books/about/Commercial_arithmetic.html?id=CCBRAAAAYAAJ&redir_esc=y > [Accessed 31 August 2012].Krisch, A• ., An Analysis of commercial Arithmetic, [Online] Available at: < http://www.springerlink.com/content/k8537j674x3w0031/ > [Accessed 31 August 2012].Kumar, S., 2008, • Polynomial Interpolation, [Video Online] Available at: < http://www.youtube.com/watch?v=Oy3uudRXolE > [Accessed 31 August 2012].Kumar, S., • Numerical methods and programming, [Video Online] Available at: < http://www.youtube.com/watch?v=ZG_TgdyDrf0 > [Accessed 31 August 2012].

Recommended ReadingMoore, J. H., 2008. • New Commercial Arithmetic, Bibliobazaar LLC.Morgan, A. D., 1900. • Elements of Arithmetic, Taylor and Walton.Calder, F., 1852. • Elementary Rules of Arithmetic.

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Self Assessment The price at which the goods are sold is known as _______.1.

selling pricea. cost priceb. marked pricec. net priced.

The price at which the goods are bought is known as ______.2. selling pricea. cost priceb. marked pricec. net priced.

The price at which the goods are intended to be sold is known as ___________ .3. selling pricea. cost priceb. marked pricec. net priced.

When the Selling price is lesser than the marked price ,it is said that the seller has given a ______ .4. profita. discountb. lossc. gaind.

If the selling price is less than the cost price ,then it is said the seller has made a _____.5. profita. discountb. lossc. gaind.

Amansellsapenataprofitof20%.Hadheboughtitat20%lessandsolditforRs5less,hewouldhavegained6. 25% .What is the CP of the pen?

20a. 22b. 23c. 25d.

If the SP of 10 articles is the same as the CP of 11 articles, what is the gain percent?7. 7a. 10b. 9c. 8d.

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What will be the simple interest on Rs 625 at 6.5% per annum for 2.5 years?8. 101.56a. 102.56b. 103.56c. 104.56d.

What is the compound interest on Rs 4000 for 9 months at 6% per annum,the interest being reckoned 9. Quarterly?

181.70a. 182b. 182.71c. 183d.

_________ is a fees paid by a borrower of assets to the owner of the assets as form of compensation for the 10. use of assets.

SIa. CIb. Recurring c. Interest d.

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Chapter IX

Relations and Functions

Aim

The aim of this chapter is to:

definefunction•

introduce the mathematical concept of relation•

highlight the domain of a relation•

Objectives

The objectives of this chapter are to:

explain the range of a relation•

elucidate the concept of range, image and co-domain•

explicate the break even analysis•

Learning outcome

At the end of this chapter, you will be able to:

understand the formula for calculating the total cost•

comprehend different ways to write a relation•

undersatnd the functi• on notation

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9.1 RelationA relation is just a set of ordered pairs. There is absolutely nothing special at all about the numbers that are in a relation.

In other words, any bunch of numbers is a relation so long as these numbers come in pairs. In maths Relation is just a set of ordered pairs.

Note: is the symbol for ‘SET’. Example: (0, 1), (55, 22), (3,-50)

9.2 Domain and Range of a RelationThe Domain isthesetofallthefirstnumbersoftheorderedpairs.In other words, the domain is all of the x-values.The Range is the set of the second numbers in each pair, or the y-values.

Example: if Relation is (0, 1), (55, 22), (3,-50), thenDomain is 0 55 3 Range is 1 22 -50NOTE: when writing the domain and range, do not repeat the values Relation can be written in several ways;

Ordered Pairs•Table•Graph/mapping.•

Examples:What is the domain and range of the following relation?(-1, 2), (2, 51), (1, 3), (8, 22), (9, 51)

Ans:Domain: -1, 2, 1, 8, 9Range: 2, 51, 3, 22, 51

What is the domain and range of the following relation?(-5,6), (21, -51), (11, 93), (81, 202), (19, 51)

Ans:Domain: -5, 21, 11, 81, 19Range: 6, -51, 93, 202, 51

9.3 FunctionsA function is a relationship between two sets of numbers.

We may think of this as a mapping; a function maps a number in one set to a number in another set. Notice that a function maps values to one and only one value.

Two values in one set could map to one value, but one value must never map to two values: that would be a relation, not a function.

Example

Ifwewrite(define)afunctionas:f(x) = x2 then we say: ‘f of x equals x squared’ and we have,f( - 1) = 1 f(1) = 1 f(7) = 49

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f(1 / 2) = 1 / 4f(4) = 16 and so on.

9.3.1 Range, image, co-domainIf D is a set, we can say,

, which forms a new set, called the range of f.

D is called the domain of f, and represents all values that f takes. In general, the range of f is usually a subset of a larger set.

This set is known as the co-domain of a function.

Example: With the function f(x) = cos x, the range of f is [-1, 1], but the co-domain is the set of real numbers.

NotationsWhen we have a function f, with domain D and range R, we write: If we say that, for instance, x is mapped

to x2, we also can add

Notice that we can have a function that maps a point (x, y) to a real number, or some other function of two variables.

We have a set of ordered pairs as the domain.

Recall from set theory that this is defined by the Cartesian product. If we wish to represent a set of all real-valued ordered pairs we can take the Cartesian product of the real numbers with itself to obtain

.

When we have a set of n-tuples as part of the domain, we say that the function is n-ary (for numbers n=1,2 we sayunary, and binary respectively).

9.4 Break Even AnalysisBreak-even analysis is a technique widely used by production management and management accountants.•It is based on categorising production costs between those which are “variable” (costs that change when the •productionoutputchanges)andthosethatare“fixed”(costsnotdirectlyrelatedtothevolumeofproduction).Totalvariableandfixedcostsarecomparedwithsalesrevenueinordertodeterminethelevelofsalesvolume,sales•valueorproduction atwhich thebusinessmakes neithera profitnora loss (the “break-evenpoint”).In its simplest form, the break-even chart is a graphical representation of costs at various levels of activity shown •on the same chart as the variation of income (or sales, revenue) with the same variation in activity.Thepointatwhichneitherprofitnorlossismadeisknownasthe“break-evenpoint”andisrepresentedonthe•chart below by the intersection of the two lines:

NOTE: The Break Even point is the point where the revenue from sales is equal to the cost of production.

For Calculating Total cost, we should know that,Profit(P)=Revenue(R)–Cost(C).Where,Total Cost = Fixed cost + Variable Cost.

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SummaryA relation is just a set of ordered pairs. There is absolutely nothing special at all about the numbers that are • in a relation.In other words, any bunch of numbers is a relation so long as these numbers come in pairs.•The Domain• is thesetofall thefirstnumbersof theorderedpairs,and theRange is the set of the second numbers in each pair, or the y-values.A function• is a relationship between two sets of numbers.Two values in one set could map to one value, but one value must never• map to two values: that would be a relation, not a function.Break-even analysis is a technique widely used by production management and management accountants.•

ReferencesJain, T. R., • Quantitative Methods, 2nd ed., FK Publication.Waters, D., 2006. • Quantitative Methods for Business, 4th ed., Prentice Hall Publication.Author Stream, • Relation and Functions [Online] Available at: <www.authorstream.com/.../sadamava-373982-2-1-relations-functions-ppt-relationsfunctions-powerpoint-education• />. [Accessed 31August 2012].Prof.Kamla, • Discrete Mathematics, [Video Online] Available at: < http://www.youtube.com/watch?v=7cTWea9YAJE > [Accessed 31 August 2012].Simonson, S., • Discrete Mathematics, [Video Online] Available at: < http://www.youtube.com/watch?v=_FG9hhiZipo > [Accessed 31 August 2012].

Recommended ReadingWaters, D., 2006. • Quantitative Methods for business, 4th ed., Prentice Hall Publication.Bedward, D., 1999. • Quantitative methods, Elsevier.Slater, J, C., 2007. • Quantitative Methods, Thomson Learning.

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Self AssessmentWhat is the domain of the following relation?1. (-1, 2), (2, 51), (1, 3), (8, 22), (9, 51)

-1,2,1,8,9a. 2,51,3,22,51b. -1,2,51,1,3c. 1,3,8,22,9d.

What is the range of the following relation?2. (-5, 6), (21, -51), (11, 93), (81, 202), (19, 51)

-5,21,11,81,19a. 6,-51,93,202,51b. -1,2,51,1,3c. 1,3,8,22,9d.

Which3. relations below are functions?Relation #1 (-1, 2), (-4, 40), (1, 2), (8,-51)a. Relation #2 (13, 14), (13, 5) , (16,7), (18,13)b. Relation #3 (3, 90), (4, 54), (6, 71), (8, 90)c. Relaton #4 (1,3)(2,4)(2,5)(3,4)d.

Which4. relations below are functions? Relation #1 (3,4), (4,5), (6,7), (8,9) a. Relation #2 (3,4), (4,5), (6,7), (3,9) b. Relation #3 (0,4), (4,-5), (0,0), (8,9)c. Relation #4 (8, 11), (34,5), (6,17), (8,19)d.

Which5. relations below are functions? Relation #1 (3,4), (4,5), (6,7), (3,-9) a. Relation #2 (3,4), (4,5), (6,7), (5,4) b. Relation #3 (0,4), (4,-5), (0,0), (8,9)c. Relation #4 (8, 11), (34,5), (6,17), (6,19)d.

For the following relation to be a function, 6. X cannot be what values?(8, 11), (34,5), (6,17), (X ,22)

a. 8, 34, 6

b. 11, 5, 17

c. 8, 34, 6, 11, 5, 17, 22

d. 8, 34, 6, 11, 5, 17

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For the following relation to be a function, 7. X cannot be what values?(12,14), (13,5) , (-2,7), (X,13)

12, 13, -2 b. 14, a. 5,7,13b. c. 12, 13,-2, 14, 5,7,13 c. d. 12,13,-2,14,5,7d.

For the following relation to be a function, 8. X cannot be what values?(12, 13), (-11, 22), (33, 101), (X ,22)

12, -11, 33 a. 13, 22, 101b. 12, -11, 33, 13, 22, 101, 22c. 12, 33d.

Suppose the weights of four students are shown in the following table.9 .

Student 1 2 3 4

Weight 120 100 150 130

Find domain1, 2, 3, 4a. 120,100,150,130b. 2,3,4c. 1,3,5d.

10.

udent 1 2 3 4

Weight 200 190 100 100

Find Range.200,190,100,100a. 200,100b. 200.190c. 200.190.100d.

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Chapter X

Statistics

Aim

The aim of this chapter is to:

definestatistics•

introducetheconceptofclassification•

explain the applications of statistics•

Objectives

The objectives of this chapter are to:

explain the functions of statistics•

explicate the limitations of statistics•

elucidate the frequency distribution•

Learning outcome

At the end of this chapter, you will be able to:

understand the characteristics of statistics•

comprehendthecharacteristicsofclassification•

enlist the characteristi• csofclassification

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10.1 IntroductionStatisticscanbereferredasasubjectthatdealswithnumericalfactsandfigures.Itisthesetofmathematicaltoolsand techniques that are used to analyse data. The word statistics is said to have been derived from the German word Statistics meaning political science or from Old Italian word stato meaning state or from New Latin word status meaning of which is position or form of government or political state. Statistical analysis involves the process of collecting and analysing data and then summarising the data into a numerical form.

10.2 Definition of StatisticsThe word statistics refers either to quantitative information or to a method of dealing with quantitative information.

Therearemanydefinitionstothetermstatisticsgivenbydifferentauthorswhichareasgivenbelow:Prof.A.L.Bowley• definedstatisticsas“Numericalstatementoffactsinanydepartmentofenquiryplacedinrelation to each other.”Websterdefined• statisticsas“Theclassifiedfactsrespectingtheconditionofthepeopleinastateespeciallythosefactswhichcanbestatedinnumbersorintablesofnumbersorinanytabularorclassifiedarrangement.”

10.3 Scope and Applications of StatisticsStatistics is associated with almost all the sciences as well as social, economic and political activities. The applications of statistics are so numerous and it is of great use to human beings in many ways. Science has become so important today that hardly any science exists independent of this and hence the statement-“Science without Statistics bear no fruit; Statistics without Science has no root.”

Statistical data and statistical methods are helpful in proper understanding of the economic problems and help •in solving a variety of economic problems such as wages, prices, analysis of time series etc. Statistical methods help in formulating economic policies and in evaluating their effectStatisticalmethodsarebeingwidelyusedinallbusinessandtradeactivitieslikeproduction,financialanalysis,•distribution, costing, market research, man power planning, business forecasting etc. Business executives and managers rely mainly on statistical techniques to study the need and desire of the consumers.Inindustry,statisticsiswidelyusedin‘qualitycontrol’.Tofindwhethertheproductisconfirmingtospecifications•or not, statistical tools like inspection plans, control charts etc are of great useA government’s administrative system is fully dependent on production statistics, income statistics, labour •statistics, economic indices of cost, price etc. All the departments of a government depend upon statistics for efficientfunctioningIn biology, medicine and agriculture, statistical methods are applied in the study of growth of plant, movement •offishpopulationintheocean,migrationofbirds,effectofnewlyinventedmedicines,theoriesofheredity,estimation of yield of crop, effect of fertiliser on yield, birth rate, death rate, population growth, growth of bacteria etc

10.4 Characteristics of StatisticsSome of its important characteristics are given below:•

Statistics are aggregates of facts Statistics are numerically expressed Statistics are affected to a marked extent by multiplicity of causes.

Statistics are enumerated or estimated according to a reasonable standard of accuracy.•Statistics are collected for a predetermine purpose Statistics are collected in a systemic manner Statistics must be comparable to each other

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10.5 Functions of StatisticsThe various functions of statistics are as given below:

Itsimplifiesthemassofdata.Withthehelpofstatisticalmethods,thecomplexdataissimplifiedintodiagrammatic•and graphical representations averages etc.It presents the facts in a definite form. Facts that are expressed in numbers are more convincing than expressed •instatements.Statisticshelpstopresentthedataorfactsinpreciseanddefiniteformforeasyunderstanding.It helps in comparison of data of same kind.•Statistical methods are extremely helpful in formulating and testing hypothesis and developing new theories.•It helps to predict future trends and to estimate any value of the population from the sample chosen.•It helps in bringing out the hidden relations between variables.•With the help of statistics, decision making process becomes easier.•

10.6 Limitations of StatisticsStatistics,inspiteofbeingwidelyusedinmanyfieldsandbeinginvolvedineverysphereofhumanactivity,facescertain limitations which are as follows:

Statistics does not deal with qualitative aspects like honesty, intelligence etc. It deals with only quantitative •data.It does not study individual facts because individual items taken separately do not form a statistical data.•Statistical methods can be applied only to the aggregate of facts.•Statistical tools do not provide the best solution to problems under all circumstances. It is one of the methods •of studying a problem and it should be supplemented by some other methods.Statistical analysis is based on probability and not on certainty. So statistical results are not universally true and •they are true only on an average.Common man cannot handle statistics properly, only statisticians can handle statistics properly.•The most important limitation of statistics is that they are liable to be misused and misinterpreted. Increasing •misuse of statistics has led to increasing distrust in statistics.

10.7 ClassificationClassificationreferstogroupingofdataintohomogeneousclassesandcategories.Agrouporaclasscategoryhasto be determined on the basis of the nature of the data and the purpose for which it is going to be used.

10.8 Objectives of ClassificationTo condense the mass of data: Statistical data collected during the course of an investigation is in the raw form. •Withrawdatawecan’tmakeanyconclusionunlessitisproperlyclassifiedintosmallgroupsorclasses.Topreparethedatafortabulation:Onlyclassifieddatacanbepresentedinthetabularform.•To study the relationships: Relationship between the variables can be established only after the various •characteristicsofthedatahavebeenknown,whichispossibleonlythroughclassification.Tofacilitatecomparison:Classificationhelpsustofindconclusionsbasedonthecomparisonofvariables.•

10.9 Characteristics of ClassificationThefollowingaregeneralguidingprinciplesforagoodclassification.

Exhaustive:Classificationmustbeexhaustive.i.e.eachandeveryiteminthedatamustbelongtooneofthe•classes.Mutuallyexclusive:Eachitemofinformationshouldfitonlyinoneclass,i.e.overlappingofitemsisnot•allowed.

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Suitability:Theclassificationshouldconformtotheobjectofinquiry.Forexample,ifthestudyisregardingthe•economicconditionofworkersthenclassificationmustnotbedoneonthebasisoftheirreligion.Homogeneity:Theitemsincludedineachclassmustbehomogeneous;elsethereshouldbefurtherclassification•in to sub groups.Flexibility:Agoodclassificationshouldbeflexible.Itshouldbeadjustable.Tothenewandchangedsituations•and conditions.Stability:Thebasicprincipleofclassificationshouldberetainedthroughout.•

10.10 Frequency DistributionA classificationaccording to the number possessing the same value of the Variable is known as frequency distribution of the given raw data.

Tally Marks (|): It facilitates counting the frequency of a value of a variate in a systematic manner. The distinct values of the variate are written down in ascending or descending order in a column. As we go through the given raw data, one by one a tally mark is inserted in each case against the respective value. It will be easy to count if tallymarksarearrangedinblocksoffivei.e.everyfifthtallymarkismarkedbyaslantinglineovertheprecedingfour. For example for the value of variate 5 we can give tally marks as ||||, for the value of variable 13 we can give tally marks as|||| |||||||.

10.10.1 Discrete or Ungrouped Frequency DistributionThe ungrouped frequency distribution is quite handy if the values of the variables are largely repeated otherwise there is hardly any condensation.

10.10.2 Continuous or Grouped Frequency DistributionIn this form of distribution the frequencies refer to groups of values. This becomes necessary in the case of some variables which can take any fractional value and in whose case an exact measurement is not possible. e.g. the height, Weight income ,etc.

10.10.3 Cumulative Frequency DistributionIn cumulative distribution, the cumulative frequencies (c.f.) are derived by successively adding the frequencies of the successive individual class intervals. The cumulative frequency of a given class can be represented by the total of all the previous class frequencies including the frequency of that class. There are two types of cumulative frequencies.

‘less than’ type: It will represent the total frequency of all classes less than and equal to the class value to which •it relates.‘more than’ type: It will represent the total frequency of classes more than and equal to the class value to which •it relates.

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SummaryStatisticshelpsincreatingmoreefficiencyinthedecisionmakingprocess•Statistics can be said as a collection of methods for planning experiments, obtaining data, and then organising, •summarising, presenting, analysing, interpreting, and drawing conclusionsThe purpose of statistics is to obtain some overall understanding of group characteristics•It is important to know how to understand statistics so that improper judgments are not made•Statistics does not deal with qualitative aspects like honesty, intelligence etc. It deals with only quantitative •data.It does not study individual facts because individual items taken separately do not form a statistical data.•Exhaustive:Classificationmustbeexhaustive.i.e.eachandeveryiteminthedatamustbelongtooneofthe•classes.Homogeneity:Theitemsincludedineachclassmustbehomogeneous;elsethereshouldbefurtherclassification•in to sub groups.Flexibility:Agoodclassificationshouldbeflexible.Itshouldbeadjustable.Tothenewandchangedsituations•and conditions.Stability:Thebasicprincipleofclassificationshouldberetainedthroughout.•

ReferencesSharma, J. K., 2009. • Business Statistics, 4th ed., Dorling Kindersley Pvt. Ltd.Jain, T. R. and Agarwal, S. C., 2009.10. • Statistics for BBA, Statistics, VK Enterprises.Medhi, J., 2005. • Statistical Methods, Methods of Data Collection, 1st ed., New Age International Publishers, pp8-12.Rajagopalan, S. P. and Sattanathan, R., 2008. • Business Statistics and Operations Research, Tata McGraw-Hill Education, pp 1-6.2012,• Statistics, [Video Online] Available at: < http://www.youtube.com/watch?v=CWrgEjXGzfg > [Accessed 31 August 2012].Judge, D• ., Statistics, lecture 1, [Video Online] Available at: < http://www.youtube.com/watch?v=24jvU95WCtQ > [Accessed 31 August 2012].Leidn university • Introduction Probability and Statistics [PDF] Available at: <http://www.math.leidenuniv. nl/~redig/lecturenotesstatistics.pdf>. [Accessed 14 July 2012].Star, Statistics http://www.stat-help.com/intro.pdf , [Accessed 14 October 2010].•

Recommended ReadingJain, T. R. and Agarwal, S. C., 2009. • Statistics for BBA, Statistics, VK Enterprises.Medhi, J., 2005. • Statistical Methods, Methods of Data Collection, 1st ed., New Age International Publishers.Rajagopalan, S. P. and Sattanathan, R., 2008. • Business Statistics and Operations Research, Tata McGraw-Hill Education.

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Self AssessmentStatistics deals with which of the following?1.

Qualitative dataa. Qualitative and quantitative data b. Quantitative datac. Productive datad.

Which of the following is not a characteristic of Statistics?2. Statistics are aggregates of facts.a. Statistics are affected to a marked extent by multiplicity of causes.b. It helps in bringing out the hidden relations between variables.c. Statistics are collected in a systemic manner.d.

Common man cannot handle statistics properly, only 3. can handle statistics properly.techniciansa. statisticiansb. artisansc. administratorsd.

“Bystatisticswemeanquantitativedataaffectedtoamarketextentbymultiplicityofcauses.”Thisdefinition4. isdefinedby:

Yule and Kendalla. Websterb. Prof.A.L.Bowleyc. Tippetd.

Which among the following is not a function of statistics?5. Itpresentsthefactsinadefinitefor.a. With the help of statistics, decision making process becomes easier. b. Itsimplifiesthemassofdata.c. It helps in comparison of data of different kind.d.

Statistics can be referred as a subject that deals with 6. factsandfigures.Alphanumerical a. Alphabeticalb. Numericalc. Quantitatived.

“Theclassifiedfactsrespectingtheconditionofthepeopleinastateespeciallythosefactswhichcanbestated7. innumbersor in tablesofnumbersor inanytabularorclassifiedarrangement.”Thisdefinition isgivenby__________.

Tippeta. Peter Drucker b. Webstarc. Oxfordd.

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_______of data leads to false conclusions.8. Misinterpretationa. Understandingb. Conceptualisationc. Summarising d.

Which statement is true?9. Statistical analysis is based on certainty and not on probability.a. Common man can handle statistics properly.b. Statistical tools provide the best solution to problems under all circumstances.c. Statistics does not study individual facts because individual items taken separately do not form a statistical d. data.

Discrete distribution is also called _______________ .10. Cumulative frequency distribution a. Grouped frequency distributionb. Ungrouped frequency distributionc. Continuous distributiond.

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Application I

Probability

When trading stocks it is of paramount importance to remove phrases like “I hope, I wish, if only” from your vocabulary.Insteaditdealswithstocksinprobabilities.Whatistheprobabilityofprofitingfromthisstockifxequals 1 and y equals 2. By calculating probabilities two things happen. All emotion is removed from trading and a dispassionate approach to trading can take place and we can better assess the risk to reward ratio and determine if this is a trade that we want to undertake. In trading, emotion is the bullet that will kill you. Stick with probabilities.

It as been owned a sizable stake in a company called General Moly which will use as a proxy for my thesis. When ithasbeenfirstlearnedofithasbeenstudiedtheprojectanddeterminedthatitownedtherightstotheMt.Hopeprojectwhichisthecompany’sflagshipproject.Ithasbeenlearnedthatwhenitreceivedthepermittobeginbuildingthe infrastructure for the mine that it would be sitting on the world’s largest deposit of molybdenum. It has also been learned that it owned a site called the Liberty project that it would deal with at a later date, but from the preliminary analysis it had done the company is sure that this project will also produce a windfall amount of molybdenum.

So what’s so great about molybdenum? Molybdenum is an alloy that is a by-product of mining copper. When molybdenum is mixed with iron ore one can create lightweight, incredibly strong stainless steel that is impervious to high degrees of temperature and pressure. It has been concluded that this would make it an essential ingredient for building oil rigs and nuclear reactors as well as basic infrastructure. It has also been learned that while China exported 97% of the world’s rare earth elements (that has subsequently changed) the country actually was a net importer of molybdenum.

Mynextstepwastolearnaboutthemanagementteam.Ithasbeenreadeverythingcouldfindonthemanagementof this company and was very pleased to learn that CEO Bruce Hansen had amassed an amazing team that was not looking to be acquired but instead looking to take this company into the arena of being the suppliers of molybdenum to the world.

What did the balance sheet look like? The company had received a loan from one of the largest producers of steel intheworld,aSouthKoreancompanycalledPosco.IthadalsoreceivedfinancingfromaJapanesecompanycalledSojitzandfinallyithadreceivedabridgeloanfromaChinesecompanycalledHanlongfortheamountof$50millionwiththepromisethatwhenpermitteditwouldreceiveanadditional$665millionloantofasttracktheprojectintoproduction. Make no mistake; the South Koreans, the Japanese and the Chinese were not being benevolent. They do not want the loans paid back in dollars. They want the loans paid back in molybdenum.

So everything was going as scripted when a “Black Swan” appeared out of nowhere and in December of 2010 with thestocktradingataround$7.00anissuecameupregardingwaterrights.TothecreditofGeneralMolyithasfromthe beginning of the project dealt with every problem with complete transparency. It has followed the law “chapter and verse” and there have been no mistakes made. This was no different. This did, however, slow the process down and the stock immediately became the backyard of the short sellers who used this as an opportunity to further drive the price down.

We will fast forward to last week when the state engineer granted the water rights to General Moly. I expected the stock to bounce to its previous level before the water rights problem but that did not happen. What had changed fundamentally?Igotoutmypencilandsharpenedit.TousethelatestbalancesheetfiguresIcouldfind,IdeterminedthatifIusedaproxypriceof$15.00albformolybdenumthenthecompanyshouldhaveanetpresentvalue(NPR)of$1.2billionora66%increaseoverwherethestockispresentlytradingasfarasmarketcapitalization.Giventhediscount to the NPV, does this imply that the market believes that there is a 66% chance that this project will not go into production? I conclude that this is simply not the case.

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Let’s honestly look at the risks at this point. Let’s assume the worst case scenario, China doesn’t come through with theexpectedfinancingortheRecordofDecision(ROD)comesbacknegativeortheMolybdenumconcentrationsaresignificantlylessthanexpectedorfinallymanagementcannotperformadequatelyenoughtogetthisthroughtoproduction.China is contractuallyobligated toprovidefinancingoncepermitted andmoreover it needs themolybdenum. There have been extensive exploratory bore holes done and everyone knows what is under the ground. The management team has performed spectacularly at every turn and I see no evidence of that changing. The X factor as I see it is the ROD and every due diligence has been taken. I see this as a probability of 80% to 90% that the ROD willcomebackpositively.SootherthantheROD,howcanthiskindofdiscounttotheNPVbejustified?

Iconcludethatthestockbasedonthefundamentalanalysisshouldbetradingatthe$10to$12.00range.Indeedthepreviouslymentionedpriceofmolybdenumthatmyanalysiswasbasedonwas$15.00apoundandasIwritethis,Freeport Mac Moran, one of the largest copper producers in the world, reported that it had sold the molybdenum (whichIsaidwasabyproductofcoppermining)atapriceof$18.00perpound. Ibelieve this ishedgefundscontinuing to try and drive the price down and judging by what I see on the tape they are beginning to run out of gas.Yesterdaythestockclosedat$4.78butIdonotseeittradingthereformuchlonger.SoIwouldencourageanyonewhoisinterestedinmakingaprofittotakealookatthisprojectandifyoulikewhatyouseestakeyourclaim at a very discounted price.

I have learned through bitter experience to “never say never.” But as I have previously stated I see the probability of this project coming to fruition at 80% to 90%. Surely somebody in Nevada wants a billion dollar mine to operate there. Consider the tax revenues generated for the state and the town of Eureka and the thousands of good paying jobs this project will create.

Source: Available at :< http://sekhon.berkeley.edu/papers/QualityQuantity.pdf> [Accessed 03 September 2012].

Questions:What did the balance sheet look like?1. AnswerThe company had received a loan from one of the largest producers of steel in the world, a South Korean company calledPosco.IthadalsoreceivedfinancingfromaJapanesecompanycalledSojitzandfinallyithadreceivedabridgeloanfromaChinesecompanycalledHanlongfortheamountof$50millionwiththepromisethatwhenpermitteditwouldreceiveanadditional$665millionloantofasttracktheprojectintoproduction.

What is great about molybdenum?2. AnswersMolybdenum is an alloy that is a by-product of mining copper. When molybdenum is mixed with iron ore one can create lightweight, incredibly strong stainless steel that is impervious to high degrees of temperature and pressure.

At what range the stock should be traded?3. Answerthestockbasedonthefundamentalanalysisshouldbetradingatthe$10to$12.00range.

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Application II

PERMUTATIONS

ThisintroducesamodifiedParticleSwarm.Optimizerwhichdealswithpermutationproblems.Particlesaredefinedaspermutationsofagroupofuniquevalues.Velocityupdatesareredefinedbasedonthesimilarityoftwo particles. Particleschangetheirpermutationswitharandomratedefinedbytheirvelocities.Amutationfactorisintroducedto prevent the current Best from becoming stuck at local minima. Preliminary study on the n-queens problem shows thatthemodifiedPSOispromisinginsolvingconstraintsatisficationproblems.

A permutation problem is a constraint satisfaction problem with the same number of variables as values, in which each variable takes a unique value. Any solution can be thought of as assigning a permutation to the variables. When apermutationsatisfiesalltheconstraints,itisconsideredafeasiblesolution.Forapermutationproblem,theremightbe one or multiple feasible solutions. The n-queens problem is one of the best examples of permutation problems. Permutation optimisation problems have been found in many areas. There are many techniques developed to handle permutation problems. In this paper, a new method called particle swarm optimization (PSO) is introduced to handle the permutation problems.

The n-queens problem consists of placing n queens on an N by N chess board, so that they do not attack each other, i.e. on every row, column or diagonal, there is only one queen exists. It is a classical complex constraint satisfaction problemin theartificial intelligence(AI) area. It has been used as a benchmark for developing new AI search techniques. During the last three decades, the problem has served as an example and benchmark for backtracking algorithms, permutation generation, the divide and conquer paradigm, constraint satisfaction problems, neural networks, and genetic algorithms. Also, the n-queens problem has many practical applications such as VLSl testing, airtrafficcontrolmodem communication systems, message routing, load balancing in multiprocessor computers, data compression, computer task scheduling, and optical parallel processing [I].

Then-queensproblembasthreevariants:findingonesolution,findingafamilyofsolutions,andfindingallsolutions.Thispaperdealswithfindingonesolutionwithinafamily.

PSO is a population based stochastic optimisation technique developed by Eberhart and Kennedy in 1995, inspired bysocialbehaviorofbirdflockingorfishschooling. During the past several years, PSO has been successfully appliedtomultidimensionaloptimizationproblemsartificialneuralnetworktrainingmultiobjectiveoptimisationproblems. However, there is no research on permutation optimisation reported in the literature.

Similar to Genetic Algorithms (GAS), PSO is a population based optimization tool. The system is initialized with a population of random solutions and searches for optima by updating potential solutions over generations. However, unlike GA, PSO has no evolution operators such as crossover and mutation. In PSO, the potential solutions, called particles,“fly”throughtheproblemspacebyfollowingthecurrentbetter-performingparticles.Eachparticlekeepstrackofitscoordinatesintheproblemspacewhichareassociatedwiththebestsolution(fitness)it

hasachievedsofar.(Thefitnessvalueisalsostored.)Thisvalueiscalledpbest.Another“best”valuethatistrackedby the particle swarm optimizer is the best value, obtained so far by any particle in the neighborhood of the particle. This location is called nbest. When a particle takes all the population as its topological neighbors, the best value is a global best and is called gbest

Source: Available at :< http://bit.csc.lsu.edu/~jianhua/chindu.pdf> [Accessed 03 September 2012].

Questions:What does n-queen problem consists of?1. What is genetic algorithm?2. What is permutation optimization problem?3.

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Application III

Matrices and Determinants

Area of a TriangleConsider a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3). If the triangle was a right triangle, it would be prettyeasytocomputetheareaofthetrianglebyfindingone-halftheproductofthebaseandtheheight.

However, when the triangle is not a right triangle, there are a couple of other ways that the area can be found.

Heron’s FormulaIfyouknowthe lengthsof the threesidesof the triangle,youcanuseHeron’sFormula tofindtheareaof thetriangle.

In Heron’s formula, s is the semi-perimeter (one-half the perimeter of the triangle). s = 1/2 (a + b + c) Area = sqrt (s ( s-a) ( s-b) ( s-c) )Consider the triangle with vertices at (-2,2), (1,5), and (6,1).

Usingthedistanceformulas,wecanfindthatthelengthsofthesides(arbitrarilyassigninga,b,andc)area=3sqrt(2), b = sqrt(61), and c = sqrt(73).

Using those values gives us ...s = 1/2 ( 3 sqrt(2) + sqrt(61) + sqrt(73) ) s - a = 1/2 ( - 3 sqrt(2) + sqrt(61) + sqrt(73) ) s - b = 1/2 ( 3 sqrt(2) - sqrt(61) + sqrt(73) ) s - c = 1/2 ( 3 sqrt(2) + sqrt(61) - sqrt(73) )s ( s - a ) ( s - b ) ( s - c ) = 1089 / 4

When you take the square root of that, you get 33/2, so the area of that triangle is 16.5.

Problems with Heron’s Formula include Mustknowthelengthsofthesidesofthetriangle.Ifyoudon’tthenyouhavetousethedistanceformulatofind•the lengths of the sides of the triangle.You have to compute the semi-perimeter, so chances are you will have fractions to work with.•Lots of square roots are involved. For the lengths of the sides of the triangle and for the area of the triangle.•It’s not the easiest thing in the world to work with.•

Geometric TechniqueThe triangle can be enclosed in a rectangle. The vertices of the triangle will intersect the rectangle in three places, forming three right triangles. These triangles are denoted A, B, and C in the picture.

The area of the triangle we desire will be the area of the rectangle minus the areas of the three triangles.

Thelegsofthethreetrianglescanbefoundbysimplesubtractionofcoordinatesandthenusedtofindtheareasince the area of a triangle is one-half the base times the height.Area of triangle A = 3 ( 3 ) / 2 = 9/2. Area of triangle B = 5 ( 6 ) / 2 = 15. Area of triangle C = 8 ( 3 ) / 2 = 12.

The sum of the areas of the triangles is 9/2 + 15 + 12 = 63 / 2 or 31.5.

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The area of a rectangle is base times height, so the bounding rectangle has area = 8 ( 6 ) = 48.

The area of the triangle in the middle is the difference between the rectangle and the sum of the areas of the three outer triangles.

Area of triangle = 48 - 31.5 = 16.5.Web Link:

Source: Available at:< http://people.richland.edu/james/lecture/m116/matrices/applications.html > [Accessed 31 August 2012].

Questions:Whatistheformulatofindareaoftriangle?1. What are the geometric Technique?2. What does heron’s formula include?3.

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Bibliography

References2011, • Arithmetic & Geometric progressions, [Video Online] Available t: < http://www.youtube.com/watch?v=ze0hNuxJaVE > [Accessed31 August 2012]2011, • Matrices, [Video Online] Available at: < http://www.youtube.com/watch?v=9tFhs-D47Ik > [Accessed 31 August 2012].2012,• Statistics, [Video Online] Available at: < http://www.youtube.com/watch?v=CWrgEjXGzfg > [Accessed 31 August 2012].Agarwal, B. R., 2007. • Programmed Statistics, Interpolation, 2nd ed., New Age International, pp 405-425.Akekar, R., 2008.• Discrete Mathematics: Set theory, 2nd ed., Dorling Kindersley Publication India, pp109-123.Arithmetic and geometric progressions• , [Online] Available at: < http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-apgp-2009-1.pdf > [Accessed 31 August 2012].Arithmetic and geometric progressions,• [Online] Available at: < http://maths.mq.edu.au/numeracy/web_mums/module3/Worksheet36/module3.pdf > [Accessed 31 August 2012].Bartlett, A., • Simple Mathematical Logic, [Video Online] Available at: < http://www.youtube.com/watch?v=PlSfS5-jU_g > [Accessed 31 August 2012].Bourke, P., • Interpolation methods, [Online] Available at: < http://paulbourke.net/miscellaneous/interpolation/ > [Accessed 31 August 2012].Dr. Kala, V. N. and Rana, R., 2009. • Matrices, 1st ed., Laxmi Publication ltd.Dr. Kirthivasan, K., • Propositional Logic, [Video Online] Available at: < http://www.youtube.com/watch?v=xlUFkMKSB3Y > [Accessed 31 August 2012].Fill, J., • Journal of theoretical Probability, [Online] Available at: <http://www.springer.com/mathematics/probability/journal/10959> [Accessed 31 August 2012].Fulda, J. S., 1993. • Exclusive Disjunction and the Bi-conditional: An Even-Odd Relationship, Mathematics Magazine.Grinstead, C. M. and Snell, J. L., 1997. • Introduction of Probability: Probability, AMS Bookstore, pp133-137.Gunawarden, J., • Matrix algebras for beginning, [Online] Available at: < http://vcp.med.harvard.edu/papers/matrices-1.pdf > [Accessed 31 August 2012].Guy, R. K., 1994. • Unsolved Problems in Number Theory, 2nd ed., Springer-Verlag, pp. 15-18.Hallie, P. P., 1954. • A Note on Logical Connectives, Mind 63.Hardy, G. H. and Wright, E. M., 1979. • An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, New York.Hurst, W., • Matrices & determinants, [Video Online] Available at: < http://www.youtube.com/watch?v=havr-W8IwKs > [Accessed 31 August 2012].IIT JEE, • Permutation and combination, [Video Online] Available at: < http://www.youtube.com/watch?v=b8mvJm1vvdQ> [Accessed 31 August 2012].IIT JEE, • Permutation and combination, [Video Online] Available at: < http://www.youtube.com/watch?v=b8mvJm1vvdQ> [Accessed 31 August 2012].Interpolation• , [Online] Available at: < http://www.mathworks.com/moler/interp.pdf > [Accessed 31 August 2012].Jain, T. R. and Agarwal, S. C., 2009.10. • Statistics for BBA, Statistics, VK Enterprises.Jain, T. R. and Aggarwal, S. C.,2010. • Business Mathematics and Statistics, V.K Enterprises.Jain, T. R.. and Sandhu, A. S., 2006-07, • Quantitative Methods: Interpolation, VK Publication, pp 7.1-7.38.

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Simonson, S., • Discrete Mathematics, [Video Online] Available at: < http://www.youtube.com/watch?v=_FG9hhiZipo > [Accessed 31 August 2012].Simpson, S., • Mathematical Logic, [Online] Available at: < http://www.math.psu.edu/simpson/courses/math557/ > [Accessed 31 August 2012].T. • Veeraranjan, 2008. Discrete Mathematics with graph theory and Combinatorics: Set theory, 7th ed., McGraw- Hill Publication, pp 51-64.

Recommended ReadingAnton, H., 2010. • Elementary Linear Algebra, 10th ed., FM Publications.Bedward, D., 1999. • Quantitative methods, Set theory, Elsevier.Dean McCullough, P., 1971. • Logical Connectives for Intuitionist Propositional Logic, Journal of Symbolic Logic.Greub, W., 1975. • Linear Algebra graduate texts in mathematics, Springer.Hallie, P. P., 1954. • A Note on Logical Connectives, Mind 63.McMahon, D., 2005. • LinearAlgebraDemystified, McGraw-hill publication.Slater, J, C., 2007. • Quantitative Methods, Set theory, Thomson Learning.Wansing, H., 2006. • Logical Connectives for Constructive Modal Logic.Waters, D., 2006. Quantitative Methods for business, Set Theory, 4th ed., Prentice Hall Publication.•

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Self Assessment Answers

Chapter Ic1. d2. b3. a4. b5. b6. a7. c8. d9. b10.

Chapter IIa1. b2. c3. a4. b5. d6. a7. b8. b9. b10.

Chapter IIIa1. c2. a3. b4. c5. a6. a7. d8. a9. b10.

Chapter IVa1. d2. b3. b4. a5. c6. d7. b8. c9. a10.

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Chapter Va1. b2. b3. b4. a5. d6. c7. c8. c9. d10.

Chapter VIa1. b2. a3. b4. d5. a6. d7. d8. b9. c10.

Chapter VIIa1. b2. a3. d4. c5. a6. a7. d8. c9. a10.

Chapter VIIIa1. b2. c3. b4. c5. d6. b7. a8. c9. d10.

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Chapter IXa1. b2. c3. a4. b5. a6. a7. d8. a9. b10.

Chapter Xc1. c2. b3. a4. d5. c6. c7. a8. d9. c10.