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TABLE OF CONTENTS

1.00 INTRO DIFFERENTIAL CALCULUS.

1.10 The limiting value of a function

1.20 The slope of a tangent line

1.30 The continuity of a function.

1.40 Differentiation as excremental notation.

1.50 Techniques of differentiation

1.60 The second derivative

1.70 Turning points: (stationary and inflexion)

Minimum points (stationary and inflexion)

1.80 Business applications

2.00 INTRO INTEGRAL CALCULUS

2.10 Define integration

2.20 Methods of integration

2.30 Solution of definite and indefinite integrals.

2.40 Business applications

3.00 INTRO SET THEORY, PERMUTATIONS AND

COMBINATIONS

1.10 SET Theory: Definition of a set

3.20 Representation / Notation

3.30 Elements of a set, sunsets and supersets,

cardinal number of sets.

3.40 Interrelations of sets: complement,

Intersection, union of sets.

3.60 The venus Diagram (Leonard Ender and John

venn)

3.60 Define permutation and

3.70 List Example of permutations and

combinations.

3.80 permutations with similar / identical objects

conditional permutations.

3.90 Applications of permutations and

combinations.



INTRO 4.00 BINOMIAL THEOREM

4.10 The meaning of the theorem

4.20 The pascal’s triage: features of the binomial

expansion.

4.30 Use of the combination formula in binomial

expansion.

4.40 Linear approximations and the binomial

expansion.

INTRO 5.00 THE IDEAR OF PROPERBILITY

5.10 Definition and meaning

5.20 Basic rules of probability

5.30 Conditional probability

5.40 Bayes theorem

5.50 Applications to business

INTRO 6.00 LINEAR PROGRAMING.

6.10 The meaning and use of linear programming

6.20 Assumptions of L.P. Model

6.30 Expressing L. P. problems.

6.40 Graphical solution to L. P. problems.

6.50 Simplex method for solving L.P. problems.

6.60 Advantages and disadvantage of graphs and

simplex methods.

INTRO 7.00 TRANSPORTATION MODEL

7.10 Meaning And Use Of The Model

7.20 General features of the model.



CHAPTER ONE

INTRODUCTION TO CALCULUS

Like most science subjects, mathematics can be divided into several branches,

among which are geometry, Algebra, Arithmetic measurements calculus,

probabilities and statistics and trigonometry. Each branch of subject has several

sub-branches, with each sub-branch serving specific functions. The central theme of

this chapter is the branch called ‘calculus’. Calculus is a wide and complex aspect

of Mathematics, which can be applied in several fields of human endeavour. The

scope of this chapter will however be limited and finance.

There are two main divisions of calculus: differentiation (also call integral

calculus).

Calculus is used for measuring the rate at which a quantity is changing, where

change is represented as slope, gradient or differential coefficient of the graph of

that variable in comparison with another calculus is used to find the gradient of

straight lines and curves. On the other hand, the integral calculus is a term which

describes a method for running up changing quantities. With the integral calculus, it

is possible to determine the areas and volumes of shapes bounded by curves.

Examine the following illustrations fo he descriptions above;

1. Given the Linear equation y = mx + c’

Y = total costs

X = quantity produced/sold and

M and c are constants, can be graphed thus:

Y

Mx

c

x



The initial part of the graph “C’ is held constant, irrespective of variations in the

value of x, and as such has not rate of change. However, the letter part ‘mx’ increases

as the value of x increases/decreases. Hence, the earlier part ‘c ‘ can be likened to the

fixed part of cost while the later ‘mx’ ca n be described as the variable component of

cost.

II GRADIENT

This simply means rate of change of one quantity measured against another.

Examples of useful gradiets are:

1. rate of change of velocity moved with time, also called accelerations

and;

2. rate of change of quantity against price, also called elasticity

A. Where the relationship between the two variables is linear, that is the graph

of the variables is a straight line, gradient can be found using:

M = vertical distance = y2 – y1

Horizontal distance = X2 – X1.

A gradient is also called a slope, and it must have a sign and size.

Vertical distance

Horizontal distance

Illustration by graph

Y B

Y2 Y-Y

Y A C Slope = m = gradient =y2 – y1 X2 – X1

Xx

B. Where the relationship between the variables is non-linear, this presents a moredifficult problem to solve. The statement Y2-Y1/X2-X1 is still relevant, but will only

give the gradient of the curve between two specific points. In other words, the

gradient of a curve is not the same at all intervals. The process of differentiation

provides a ready means of finding the slopes/ gradients /rates of change of such



functions and their turning points. Integration, on the other hand, helps to determinethe total amount of costs or revenue between two levels of activity.

111. THE LIMITING VALUE OF FUNCTION

Suppose that y=x2. The graph of y=x2 for values of x from o to 4 is as shown below.y

16

14 C12

10 C

8

6

4 B

2 A

0 1 2 3 4 X

Gradient between A and B = 9 – 4 = 5

3 - 2

Gradient between B and c = 16 – 9 = 54 - 3

NOW EXAMINE THE FOLLOWING GRAPH.

Y B Y2

A Y1

0 Y1 (X1+A) X

From the graph, X1 = X Y1 = X2

X2 = X + a y2 = (x +a)2

Y2 – y1 = (x + a )2 – x2

X2 – X1 = (x + a ) – x

= X2 + 2a x + a2 – x2 = 2ax + a2



x + a – x a

= 2x + a

Therefore, between A and B, gradient of AB = 2x + a. But as a tends to zero, 2x +

a = 2x. This process of calculating the limiting value of a function is calleddifferentiation.

So that since y = x2

dy = 2x, the limiting value and

dx derivative of y with respect to x.

IV SOLVING LIMITS

The solution of limits can be obtained by avoiding any one of the following

situations:

0/0, 0/z, z/0, &/&, z/&, &/zwhere z = any number (positive or negative)

& = extremely small or large number, also called infinity.

V. METHODS

Some of the methods for solving limits are listed below:

1. Direct substitution

2. Simplification

3. Factorisation

4. Rationalization

5. Differentiation and l’ hospitals Rule.

1. DIRECT SUSTITUTIONS

This should always be attempted before any other method is used. Once the

identified value are avoided, the solution of the limit is good.

EXAMPLES.

(i) line x2 + 2x + 1

x =) a. As x becomes a, x2+ 2n + 1 = a2 + 2a + 1

= ( a+ 1 ) ( a + 1 )

(ii) line x + 22x

x- ) 2

As x ) 2, x + 2 = 2 + 2 = 4 = 12x 2 (2) 4

2. Simplification and fractorisation

Examples

(i) lim x2 – 1 = lim ( x+ 1) (x – 1)

x-)1 x - 1 x-)1 (x –1)

===) lim ( x + 1)

x -) 1

:. As x-)1, x + 1 = 1 + 1 = 2



(ii) lim x( x + 2 ) (x+ 1 )x-)& x3 – x

==) lim x ( x + 2 )(x+ 1)x-)& x (x2- 1)

but x2 –1 = x2- 12 = ( x – 1 ) ( x + 1 )==) lim x ( x + 2 ) ( x + 1 ) = lim x + 2

x-)& x(x + 1 ) ( x + 1 ).X-)& x- 1

As x-) &, x + 2 = & + 2 = 2 = -2X – 1 & - 1 -1

(iii) Limi 2x3 + 4x2 + xx -) 0 x2 + 5x

Divide all through by the least power of x

==) 2x3 + 4x2 + xx x x

x2 + 5x

x x

= 2x2 + 4x + 1

x + 5

:. As x -) 0, 2x2 + 4x +1 = 2(02) + 4(0) + 1 = 1x + 5 0 + 5 5

3. Rationalization

Lim 2 - x – 3 x 2 + x – 3

x-)7 x2 – 49 2 + x - 3

= Lim 4 +2 x – 3 - 2 x - 3 - (x – 3 )x -) 7 ( x2 – 49) (2 + x – 3

= Lim 4 – x + 3 = -X + 7

X-)7 (x – 7) (x + 7) ( 2 + x – 3 (x – 7) (x+ 7 (2 + x – 3 )

= Lim -1(x –7) = Lim -1

x -) (x-7) (x +7) (2 +x-3 ) x - )7 (x+7) (2+ x – 3 )

= -1 = -1 = -1 = -1

(7+7) (2+ 7 –3) 14 (2 +2) 14 (4) 59

Differentiation ( L’hospital’s Rule)

This should be used where other methods fail, The solution obtained bydifferentiation is the same as that obtained using any other method.

EXAMPLES



(i) Lim x2 – 1x-) x – 1

Takes x2 – 1 = f (x) and x –1 = g (x)

F’ (x’) = dy/dx = 2x, g’ (x) = 1

:. Lim 2x = 2(1) = 2

(ii) Lim (x – 2) (x2 + 3x –2)

x-)2 x2 – 4

= Lim x3 + 3x2 – 2x – 2x2 – 6x + 4

x-)2 x2 – 4

= Lim x3 –x2 – -8x +4x-)2 x2-4

f’ (x) = 3x2 + 2x – 8, g, (x) = 2x

:. Lim 3x2 + 2x – 8 = 3(22) + 2 (2) – 8x-)2 2x 2(2)

= 12 + 4 – 8 = 8 = 24 4

This solution can equally be obtained thus

Lim (x – 2) (x2+ 3x – 2) = ( x2+ 3x – 2 )x-)2 (x +2 ( x – 2 ) (x + 2)

= 22 + 3 (2) – 2 = 4 + 6 - 22 + 2 4

= 8 = 24

(iii) Lim 2x3 + 4x2 + xX- )0 x2 + 5x

6’(x) = 6x2 + 8x + 1, g’ (x) = 2x + 5

:. Lim 6x2 + 8x + 1 = 6(02 ) + 8 (0) + 1 = 1x-)0 2x + 5 2(0) + 5 5

(VI) THE INCREMENTAL NOTATION.

(THE FIRST FICTION)

1. Take y = x2,Let change in y = dy and change in x = dx

:. Y + dy = (x+ dx)2

y + dy = x2 + 2xdx + dx2

dy = x2 + 2xdx + dx2 – y ( note that y = x2)

dy = x2 + 2xdx + dx2 – x2

dy = 2xdx + dx2

Dividing all through by dx,



dx = 2xdx + dx2 dx dx dx

dx = 2x + dxdx

As dx-)0, dx = 2x

dx2. Differentiation from the first principle y = x2 + 5x

y + dy = (x + dx)2 + 5(x+dx)

y+dy = x2 + 2xdx + dx2 + 5x + 5dx

y+dy = x2 + 2xdx + dx2 + 5x + 5dx

dy + x2 + 2xdx + dx2+ 5x 5dx – x2 – 5x

dy = 2xdx + dx2 + 5dx

divide through by dx

dy =2xdx + dx2 + 5dx

dx dx dx dx

dy = 2x + dx + 5

dx

As dx-)0, dy = 2x + 5

dx

dy = 2x + 5dx

3. Given y = 3x2 – 2, differentiate y with respect to x from the first principle.

Y = 3x2 – 2

Y + dy = 3(x + dx)2 – 2

Y + dy = 3(x2 +2dxdx + dx2) –2

Y + dy = 3x2 + 6xdx + 3dx2 –2

Y + dy = 3x2 + 6xdx + 3dx2 – 2 – 3x2 + 2

dy = 6xdx + 3dx2

dx = 6xdx + 3dx2 dx dx dx

dy = 6x + 3dx.dx

As dx-)0, dy = 6x

dx



4. Differentiate from the first principle y = 2x3 + 4x2 + 6x + 3

y + dy = 2(x+dx)3 + 4 ( x + dx )2 + 6 ( x + dx ) + 3

y + dy + 2(x3 + 2x2dx + xdx2 + x2dx + 2xdx2 + dx3) + 4(x2 + 2xdx + dx2) + 6x + 6dx

+ 3.

Y + dy = 2(x3 + 3x2dx + 3xdx2 + dx3) + 4x2 + 8xdx2 + 6x + 6dx + 3

Y + dy = 2x3 + 6x2 dx + 6xdx2 + 2dx3 + 4x2 + 8xdx + 4dx2 +6x + 6dx+3

dy = 2x3 + 6x2 dx + 6xdx2 + 2dx3 + 4x2 + 8xdx + 4dx2 +6x + 6dx+3

-2x3 – 4x2 – 6x –3

dy = 6x2 dx + 6xdx2 + 2dx3 + 8xdx + 4dx2

dy = 6x2 dx + 6xdx2 + 2dx3 + 8xdx + 4dx2 + 6dx

dx dx dx dx dx dx dx

dy = 6x2+ 6xdx + 2dx2 + 8x + 4dx + 6

dx

As dx -)0, dy = 6x2 + 8x + 6

5. Once again, consider y = x2, dy 2x from this,

dxwe can infer that fir every y = axn,dy/dx = anx n-1

Therefore, given y = axn,dy/dx = anxn-1

The blocked formula is considered the general principle of differentiation.

Vii TECHNIQUIES OF DIFFERENTIATIONA. Monomial and polynomial

In general, for every y = axn, dy = anxn-1

dx

EXAMPLES

1. Find dy/dx if y = mx,

dy/dx = m (mx1 x X1-1).

2. Differentiate with respect to x,

y = 6x2

dy/dx = 12x (6x2xX2-1).

Where the function is a sum, the differentiate coefficient is the sum of the

differentiate coefficients of the separate parts.



3. Given y = 2x3 + 4x2 + 6x

dy/dx = 6x2 + 8x +6

To differentiate a whole number, say 5, attack x as follows,

4. Given that y = 3x2 + 7x + 5

y = 3x2 + 7x + 5x0

dy/dx = 6x + 7 + 0

= 6x + 7 =

B PRODUCT RULE

Let u and v represent functions of x and y, such that y = uv, by implication

dy of uv = u dv + v du

dx dx dx

Examples

1. Find dy/dx if y = (2x + 3) (3x2 + 2)

Take 2x + 3 = 3 = u and 3x2 + 2 = v

Since u = 2x + 3 dx2 3x22 + 2

du = 2 dv = 6xdx dx

From dy u dv/dx + v dv/dx

dxdy = (3x2 + 2)2 + ( 2x + 3) 6x ( note: a+b = b + a).

= 6x2 + 4 + 12x2 + 18x

=18x2 + 18x + 4

= 2(9x2 + 9x + 2)

2. Find dy/dx in y = (3x – 1) (4x +6)

Take u = 3x – 1 and v = 4x + 6

du = 3 and dv = 4dx dn

:. dy = (3x – 1)4 + (4x + 6)3dx

=12x – 4 + 12x + 18

= 24x +14

= 2(12x + 7)



3. If y = (x2 + 4 ) ( 6x1/2 + 3 )

du = 2x and dv =3x-1/2

dx dx

dy = (x2 + 4) 3x-1/2 + (6x1/2 + 3) 2x.

= 3x3/2 + 12x-1/2 + 12x3/2 + 6x

= 15x3/2 + 6x + 12x-1/2

4. Find dy/dx given that y = (3x2 – 2x + 1) (4x +5)

Take 3x2 – 2x + 1 = u and 4x +5 = v

dy = v du = (4x +5) (6x – 2) + (3x2 –2x +1)4dx dx

= 24x2 – 8x +30x – 10 + 12x2 – 8x +4= 36x2 + 14x –6.

C QUOTIENT RULE

Let u represent the function of x which is the numerator and v represent that which

is the divisor.

Consequently, y = u

V

dy = v du = u dv

dx dx dx

v2

Examples

1. differentiate y = x2 - 1x3 + 1

Take x2 – 1 = u and x3 + 1 = v

du = 2x dv = 3x2

dx dx

dy = v du - u dv = (x3 + 1) 2x – (x2 – 1) (3x2)

dx dx dx (x3 + 1) (x3 + 1)v2

= 2xn + 2x – 3x4 + 3x2 = -x4 + 3x2 +2x

x6 + x3 + x3 + 1 x6 + 2x3 +1

2. Find dy if y = x - 1

dx x +1

Take u = x – 1 and v = x + 1

d du 1, dv = 1

dx dx



dy = (x + 1)1 - ( x – 1)1 = x +1 – x + 1 = 2

dx (x + 1)2 x2 + 2x + 1 x2 + 2x + 1

3. If y = 4x3 + 2, find dy

x2 dx

Take u = 4x3+ 2 and v = x2

du = 12x2, dv = 2x

dx dx

dy = x2 (12x2) – (4x3 + 2) 2x = 12x4 – 8x4 – 4x

dx (x2)2 x4

4x2 – 4x = 4x (4x3 -2) = 4(x3 – 1)

x4 x(x 3) x3

4. Given y = 3 – 2x find dy

3 + 2x dx

Take u = 3 – 2x and v = 3 + 2x

du = -2, dv = 2dx dx

dy = ( 3 + 2x ) –2 – (3 – 2x)2 = - 6- 4x – 6 + 4x 4xdx (3 + 2x)2 9 + 12x + 4x2

= -12(3 + 2x)2.

CHAPTER TWO

SET THEORY, PERCUTATIONS

AND COMBINATIONS

I. SET THEORY

A. DEFINITION



“Is a well-defined list, collection, or class of objects. The objects in sets can be

anything: numbers, people, letters, rivers, c. t. c. these objects are called the

elements or members of the set”.

……Mayo associates limited & BPP publishing limited.

“A set is a collection of things. The things have something in numbers. The

example, a set of stamps, a set of chairs, a set of numbers. The things in a set are

called its members or elements”.

…Duncan And Christine Graham, Gose Mathematics.

“ A set is simply collection of things or people, called elements.

Capital letters such as A, B, C, are used to named such collections.

…. Walter van stigh, success in mathematics.

“Mathematics call any well-defined class a set, and by well-defined we mean that

we must be able to decide definitely whether any one object does or does not belong

to that set”.

….F. G. J. Norton,. Hardwood Clarke’s ordinary level mathematics

The above definitions set out clearly the subject matter of this chapter. The

language of sets has been one form in the flesh for most students learning

mathematics or mathematics – related courses. While, the key words in the

definitions above can be cited as.

‘well-defined’ ‘collection’

‘elements or members’

B. NOTATION

Notation means representation of sets. How are sets written (represented)? Which

rule(s) must be observed in drawing sets? This

Parts provide answer to the other possible questions.

=(x2 +4)3 (33x4 + 36x2)

Generally, sets are donated by capital letters and written in

= (x2 + 4)3 (33x4 + 36x2)

inverted brackets, the elements are not arranged in any particular

= 3x2(x2 + 4)3 ( 11x2 + 12). Order.



Examine the following expert opinions on set notation:

“A set is usually named XVIII. APPLICATION PROBLEMS or labeled with a

capital letter. Curly brackets or braces () often replace the words the set of ‘. And

when members are listed the order does not matter. Commas must be placed

between members”

…Duncan and Christine graham ditto

“Capital letters such as A, B, C, are used to name such collections

by: a. describing the set in words;

i. Nis the set of al integers

ii. A is the set of all prime numbers

iii. q is the set of all past presidents of America.

B. Listed or tabulating all the elements of a set; which is only practicable if the set

is relatively small, The brackets () are used to enclose the listed elements of a set or

its description by means of a property of its elements”.

….Walter van stigt, success in mathematics.

“ sets are usually donated by capital letters

A, B, X, Y, S,……. The elements in sets are usually donated by small letters a, b, x,

y, s, ….. if A is the set of numbers 1, 2, 7, and 10, then we write A = (1, 3, 7, 10)”.

….. Mayo and BPP. Do.

From the above descriptions, a set of all positive even between o and 10 inclusive

will be written thus:

A = (2, 4, 6, 8, 10)”.

Note also that sets can be given as inequalities. Imagine a manufacturing process,

which requires its operatives to produce a minimum of 5 units but not more than 10

units of product daily. The set can be written thus:

P5 p 10

That is 5 p, otherwise written as p 5

And p 10

P = (5, 6, 7, 8, 9, 10).

In order to be able to display or decide sets which are given as inequalities, a fairly

good understanding of the following signs is essential:

‘less than’ ‘greater than’

‘less than/equal to’ ‘ greater than /equal to’



# ‘not equal to’

c. TECHNICALITIES IN NOTATION

1. The use of…………

a. Continuing a pattern to infinity.

Suppose x = (a, 2a, 3a, 4a, ……) the after the 4th member the pattern

continues indefinitely (say to infinity)

b. continuing a pattern to a limit

suppose y= (a, b, c, d,…..z) The …suggests that the pattern continues until z. thus,

the last term of the series is z.

2. The cardinal number of set.

This is simply the number of elements contained in the set Hence, if p = (5, 6, 7, 8,

9, 10,), n(f) = 6

And if y = (a, b, c, d, …..z), n(z) = 26

3.The universal set (u or e)

This is the which consists of everything, all members, all members, of a set under

examination. U or E donates it.

Suppose set A = ( all white lorries)

Set B = ( all blue lorries)

Set C = ( lories of other cal curs)

U = (all lories)

Similarly, Let X = (all primary school pupils)

Y = (all secondary school pupils)

Z = (all undergraduates)

T = ( all postgraduates students)

U = (all students).

4. SUBSETS

A subset can be defined as a smaller set within a bigger set.

It is thus a set inside a set.

A set of all even numbers is a subset of the set of all integers.



Given that A = (a, b, c, …..z) and B = (all consonants),

C = (all vowels), then Band are subsets of A, written thus:

ECA

CCA

The symbol ‘c’ means is a subset of A = all sets are subsets of their universal sets.

5. The complement of a set.

“ All the elements of the universal set which are not members a set A themselves

from a set called the complement of A, written A”.

“These set of all element s of the universal set which are not elements of set A is

called the complement of A. I is usually symbolized by a dash (‘) after the

appropriate letter”

W. V. STIGT.

From the above vices, the complement of a set is found by comparing that set with

the universal set.

Suppose set A = (Tuesday, Wednesday, Friday)

B = ( Thursday, Sunday)

U = (Monday, Tuesday,….Sunday)

Thus A = (Sunday, Monday, Thursday, Saturday)

B = (Monday, Tuesday, Wednesday, Friday, Saturday)

Similarly, given that A = (2, 4, 6, 8, 10)

B = (1, 3, 5, 7, 9,)

And U = ( 1, 2, 3, ….10)

Then A’ = (1, 3, 5, 7, 9)

B’ = (2, 4, 6, 8, 10)

6. The Union sets

A List of all the elements contained in two or more given sets is the union of the

given sets.

For example, let A = (chemistry, physics, Biology)

B = (Economics, physics, commerce)

The set of union B, written as A V B is

A V B = (Chemistry, physics, Biology, commerce, economics)



From the above, you would see that all the elements in A and B were organized and

presented as AVB.

You will also have noticed that ‘physics’ which is element of A and B, has been

written just once. This is because, by rule, when compiling the union of sets,

elements common to two or more of the sets must be written only once.

Similarly, given A = (1, 2, 4, 8) and B = (1, 3, 5, 7)

AVB = (1, 2, 3, 4, 5, 6, 7, 8)

Thus if an element can belong to the set P or to the set Q, it belongs to the union of

P and Q. This is the set of all the elements of P or Q or of P and Q.

7. THE INTERSECTION OF SETS.

Two set are said to intersect when each has element common to the other. The

intersection of two or more sets is, A n B, for sets A and B

Alternatively, if A and B are two given sets, An B = ( x/x EA, XEB).

The above needs, / intersect B is equal to x, such that x is a member of A and X is a

member of B.

Given A = (1, 2, 3, 4) and = (3, 4, 5, 6)

A n B = (3, 4)

8. EQUALITY OF SETS

Two sets are equal when they contain the same set of elements or members. The

manner (order) of arrangement of the elements does not matter.

For instance, if A = (a, c, c, z) and

B = (z, c, c, a)

A = B



D. THE VENN DIAGRAM

This is a diagrammatic representation of the interrelationship between two or more

sets and the universal set,

The diagram was invented by Leonard Enler and simplified and improved by John

Venn. It is thus named after him.

The universal set (E or LL) is represented with a rectangle (or square) while others

are represented with circles.

Now examine the following illustrations.

(i) A = (Ade, Bayo, Sola, Kola, Bode)

Ade

Bayo Kola

Sola

Bode

(ii) Given P = (1, 2, 3, 4, 5, 6, 7, 8)

U = (1, 2, 3,………..10).

9 1 3 2

6 4 5

7 8 10

(iii) suppose x = (1, 2, 4, 8)

Y = (3, 4, 6)

Z = (1, 4, 8)

X n z = (4), y n z = (4)



X Y

2 4 3

1 8 6

(iv) If the universal set for x, y and z above is u = (1, 2, 3, 4, 5,…..10)

The Venn becomes

X 2 -- 3 Y

1 2 4 --

--

Z

(V) For sets x and z

X 2 1

2 4 8

The diagram below further summarizes the interelationship of sets, using Venn

diagrams.

S/N VENN DIAGRAMS NOTATION INTERRETATION

1 . U

E or U Universal set

2. n (x) as The set of elements

Subset of u common to x and y



3.

x Y

X n Y the set of elements common to x and

Y.

4. X Y x -y The difference of sets x and y.

Also called x only.

5.

Y x - y The complement

of sets x .

6. x u y Set of all elements

in x and y or

of x or y.

7.

X’ The complement of x

X



8. A B B C A B is a subset of A

9.

P n Q = ( ) P and Q are

P Q Disjoint sets.

10.

A B AURVC The union of three

Sets A, B, and C. a

All elements in A,

B, and C.

C

11.

X Y x n y n z The union of three

Common to x, y



And z.

Z

12.

X Y (x n y n z)’ The complement

Of x n y n z. The

Elements absent

In x n y n z but

Z present in the

Universal set.

13.

X Y (x u y u z) The complement

Of x u y u z. The set of

elements absent in x u y

u z but present the

universal set.

Z

14.X Y Y n Z The intersection of Y and

z. The set of elements

common to Y and z.

Z



15.

X Y (XUY) The intersection of

The union of X and Y with Z.

The set of all elements

E. SPECIAL LAWS

THE ALGEBRA OF SETS HAS CERTAIN SPECIAL LAWS:

I. COMMUTATIVE

Generally, in binary operations, if the order of combination of, elements does not

affect results, the operation is referred to as commutative.

In a + b = b + a, addition is commutative but in a – b = b – a, subtraction is shown

no to be commutative.

For sets, A U B = B U A and

A n B = B n A.

II. ASOCIATIVE.

We can combine any three element in two ways. Suppose we intend to add a, b, and

c. Then, we either say ( a + b ) + c or a + (b + c).

Where the order of combination is immaterial, we refer to the operation as

associative.

For sets, (A U B) V C = A U ( B U C)

(A N B ) N C = A n (A n c)

III DISTRIBUTIVE

Suppose a, b and c are three integers, then a (b + c) = ab + ac. This shows that

multiplication is distributive over addition. But (b +c) :-a is not b :-a + c :-a.

Therefore division is not distributive over addition.

For sets, AU (B n C) = (A U B) n (A U C)

A n (B u c) = ( A n B) Y (A n C)



IV IDENTITY

Identity laws are general laws which affect sets,

These are:

(i) For empty sets ( null sets)

1. AU ( ) = A

2. An ( ) = ( )

3. ( ) UE = E

4. ( ) n E = ( )

5. n (A U B) = n (A) + n( B) – n(A NB).

ii For universal sets ( E or U)

1. A U E = E

2. A n E = A

3. A C E.

4. A’ U E = E

5. A’ n E = A’

6. E’ = ( )

7. ( )’ = E

8. (A’) = A

9. (E’) = E.

V DE –MORGAN’S

1. A’ n B’ = (A U B)’

2. A’ U B’ = (A n B)’

F. WORKED EXAMPLES

(a) (i) Define a set

(ii) let set s = (3,2,5,8, 16, 13) and T = ( 1, 20, 19, 5, 7, 8)

(iii) Let s = (5, 3, 1) and T = (2)

Determine s and T

(iv) Let set A = (x/x2 = 4, x is odd)

What sort of set is A

Use venn diagrams where applicable.

SOLUTION

(a) (i) A set a well – defined list, collection of objects.

(ii) S = (3, 2, 5, 8, 16 13), T = (1, 2, 0, 19, 5, 7, 8)



S n T = (5, 8)

S T

3 2 5 17

16 13 8 20

19

(iv) A = ( x/x2 = 4, x is odd)

For x2 = 4, x = + = +2

4

X cannot be odd.

:. A = ( ), an empty set.

A survey of a small town of 20000 households provided the following data.

60% have refrigerator;

35% have telephone;

15% have colour television;

18% have telephone and colour television;

12% have all three.

Using a Venn diagram,

a. How many of these households have a telephone?

b. How many have only refrigerator?

c. How many only colour television set.

d. How may have neither a telephone, nor a refrigerator, nor colour

television set?

e. What percentage of the household have only two of the three equipments?

SOLUTION

100% Of The Households = 20,000

60% = 60 X 20,000 = 12,000

100

35% = 7,000

15% = 3,000



18% = 3,600

12% = 2,400

10% = 2,000

8% = 1,000

HINT:

a. Represent each set with a letter, A, B, C, X, Y,Z.

b. Since It Is Already Established that each set has one thing or the other in

common with the next, represent each set with a circle, showing clearly their

meeting points.

c. Fill the sets, starting from the mid-point common to all.

d. Solve as required.

R T

E C F let refrigerator = R

“ Telephone = T

A “ colour T. Y = C.

C D

G

C

At point a, all three meet = 8% = 1600 households

,, ,, b, R and T meet = 13%= 10% = 200 households

,, ,, c, R and c meet = 10% - 8% = 2% = 400 households

Recall that: a + b = 18%

a + c = 10%

a + d = 12%

Similarly, a + b + d + f = n® = 60% = 12% = 12,000 households

:. C = 12,000 – (1600 + 2000 +400) = 8,00 households

Also, a + b + d + f = n(T) = 35% = 35% = 7,000 households



:. F = 7,000 – (1600 + 2000 + 800) = 2, 600 households

And a + c + d + g = n(c) = 15% = 3,000 households

:. G = 3,000 – (1600 + 400 + 800) = 200 households

THE RE-DRAWN VENN IS THUS:

R 2u 2b T From this redraw venn, we

2nc can now answer all the

2ub questions.

4n f2n

2cn

C

(a) n (Telephone) only = 2600 households

(b) n ( Refrigerator) only =8000 households

(c ) n (colour T.V) only = 200 households

(d) n ( neither T nor R no c)

= 2000/0 – n (either T or R or C)

= 200,60 – (8000 + 2000 + 1600 + 400 + 2600 +800 + 200)

= 20,000 – 15, 600 = 4, 400 households.

THE VENN CAN ALSO SHOWN IN PRECENTAGES

R T

40% 10% 13%

8%

2% 4% 22%

19% C

d. n ( % having only two) = 10% + 25 + 4% = 16%.

3. In a certain examination, 72 candidates offered Mathematics, 64 English and

62 French. 18 offered both Mathematics and English, 24 Mathematics and

French and 20 English and French. 8 offered all three subjects . How many

candidates were there for the examination?



SOLUTION

A 34

10 Let mathematics = M

8 82 English = E

E 38 26 F French = F

16

n (M n E n F) =8 n ( M n F) = 24

n (m n E) = 18 n (E n F) = 20

n( M ) = 72 n (E) = 64

n (F) = 62

n (m n F) only = 24 – 8 = 16

n ( E n F) only = 20 – 8 = 12

n (M n E) only = 18 – 8 =10

n (M) only = n (m) – n(m n E) only + n ( m n E n F)

+ n (m n F) only

n ( m) only = 72 – (10 + 8 + 16)

= 72 – 34 = 38

Similarly, for E,

N (E) only = n (E) – ( n (m n E) only + n (m n E n F) + n ( E n F) only

= 64 – (10 + 8 12)

= 64 – 30 = 34

And for F,

N ( F) only = n (F) – (n (m n F) only + n ( n F) only

+ n ( m n E n F )

= 62 – (16 + 12 + 8)

= 62 – 36 = 26

Total number of candidates for the examination

= 38 + 10 + 34 + 16 + 8 + 12 + 26

= 144 candidates.



II. PERMUTATIONS

A.DEFINITION

“ A permutation is an ordered arrangement of items. Thus AB is a different

permutation to BA even though the individual two items A and B are the same, they

are in a different order” Terry Lucy, quantitative Techniques.

“ permutation is a technical word for arrangement. A permutation is a set of items,

selected from a large collection of items, in which the order of arrangement is

significant.

…. M. A. Lawal, Business Mathematics volume Two.

“ A permutation of a n different objects with taken or at a time is an arrangement of

out of the n objects with donated by n pr p (n,r) or pn, r and is given by

npr = n(n – 1) (n-2)….(n – r +1) = n’

(n – r)’

Suppose we create boxes for the first three positions

Thus:

1st 2nd 3rd

Any one of any one of any one of

8 1 6

:. Possible ways = 8 x 7 x 6 = 336 ways.

B. THE SPECIAL.

(X’ CXM)

a product such as 5 x 4 = 3 x 2 x 1 is known as factorization written as 5’.

There were 3 x 2 x 1 = 3.

6 x 5 x 4 x 3 x 2 x 1 = 6;

Thus, n’ = n x (n – 2 ) x ( n – 3) x….

From the differentiation given by Lucy special, in America series statistics,

N m = n:

(n – r)



where in is the total number items available and is the number required or space

available.

In our first example, 3 and 4 players and need also to our n = 4 and r = 4

:. The number of arrangement is

44 = 41 = 41 = 41

(4 – 4)’ 0’ 1

note that 11 = 1 and 01 = 1

:. Nor where n = r = n’

:. 4p4 = 41 = 24 wives.

Similarly, in the second example, n = 8 and r = 3

:. The number of arrangement = 8p3 = 81

(8- 3)

= 81 = 40320 = 336 ways

5’ 128

Examples.

1. How many different numbers can be formed using the digits 2, 2, 3,

3, 3, if the numbers must all be of 5 digits?

The result is 1 , since 2 appear twice and 3 appears

2 3’ twice.

:. Number of 5 digit numbers = 120 = 10

12

2. An airline twice compare is to arrange 3 footballers, 3 boxers and 3 sprinters

on a Lufthansa flight to the International Olympic committee headquarters in

Switzerland. In how many ways can they be arranged if:

(a) each of the sportsmen must remain with their groups and;

(b) only the sprinters must sit together?

(a) Arranging each group together,

000 000 000

F B S

3’ X 3’ X 3’

6 X 6 X 6 = 216 WAYS.



But note that any one of the three groups can be arranged first = 3’

:. Number of possible arrangements = 216 x 3’ = 1296 ways.

(b) Arranging only sprinters together, this puts footballers and boxers in any

order.

That is 6’ sprinters together = 3’

Call also that either sprinters or a combination of footballers and boxers can be

arranged first. This is also 2’

:. The result is 6’ x 3’ x 2 = 8640.

D. CONDITIONAL PERMULATIIONS

The second example above has already produced this topic.

Sometimes, certain special conditions may be attached to calculating

permutation.

4. A food center offers a choice of 3 starters, 4 main courses and 2 sweets.

How many different meals are available ?

The result is 3 x 4 x 2 = 24 meals.

5. SILOT LTD has 4 training officers and 2 sections requiring their services. In

how many ways may the 4 officers be assigned to the 2 section?

Let the officers be A, B, C, D, and department p, Q,.

P can have any one of remaining 3

= 4 x3 = 12 ways.

Alternatively, number available n = 4

Number required r = 2

:. Possible assignments = npr = 4p2 = 4

(4 – 2)’

= 4 = 24 =12ways

2’ 2

5. In how many ways on the letters of the word POLYTHECHNIC be

arranged?

n= 11, r = 11

:. npr = 11 p11 = 111 = 39,916,800 ways

C. PERMUTATION WITH SIMILAR /INDENTICAL OBJECTS



Where items to arranged are not completely different, the number of

permutation will be reduced.

Suppose we want to find out the number of the ways possible to arrange

the letters of the word ONDO. The letter appears twice and you cannot

possible distinguish one from the other. If all letters were different, the result

will be 4 p4 = 4 = 24 ways.

But since appears twice, has 21 and our result becomes

4’ = 24 = 12ways

2 2

Examples

1. In how many ways may the letters of the word SCHOOL be arranged if:

(a) The two o’s must not come together and

(b) The two o’s must remain together.

(a) By excluding the o’s in school, the remaining letters S C H L can be

arranged in 4’ ways.

The first o can occupy any one of the remaining four positions vacant = 4.

Also, recall that it is not possible to distinguish between the two o’s = 2

Therefore, the number of arrangements, with the first o above can occupy.

4’ x 5 x 4 = 24 x 5 x 4 = 240 ways.

2’ 2

Examine this diagrams for the spaces which the first o above can occupy.

? s c H L

5 7 C H L

S C ? H L

S C H ? L

S C H L ?



(B) if the two os must remain together is only change in the result is that whichever

space will be occupied by the first o is automatically for both. That is, 5 spaces

:. The result is 4’ x 5 = 60 ways

2'

2. Examine the word O F F I C E R. In ho w many ways may the letters be arranged

given that:

(a)The F F should be separated;

(b) The F F should remain undetached.

(a) 0 1 c e r = 5’ ways

1st F 1 2 3 4 5 6 =6 places

1 1 1 1 1

2nd

F 1 2 3 4 5 = 5 places

For 2nd F, remember that 1st F would have occupied one out of the 6 places.

Also, the FF ar not distinguishable = 2’

Therefore, number of arrangements

= 5’ x 6 x 5 = 1800 ways

2’

(b) 0 1 C E R =5’ WAYS

F F 1 2 3 4 5 6 =6WAYS

Therefore, the result is 5’ x 6u = 360 ways

2’

III. COMBINATIONS

DEFINITION

“ There will be occasions when selections will be made where the order does not

matter meaning that a, b, is the same as b, a. This is known as a combination”.

…. T. Lucy, quantitative Techniques.

“ In combination, we are concerned with the number of different grouping of

objects that can occur without regard to their order”.



…. c..Clidi, Business mathematics, volume Two.

“.. is a technical word use in mean session. A combination is a set of items, selected

from a large section of items, regardless of the order in which they are selection.

…. M.A. Lawal, Business Mathematics, volume Two.

full hand out

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