bpa 12203 mathematics for management chapter 3

BPA 12203MATHEMATICS FOR MANAGEMENT

Chapter 3

COMBINATORICS

Mathematics for Management: Combinatorics

Factorial notation

Definition

The product of the first n natural numbers is called n factorial andis denoted by n!.

For a natural number n,

n! = n(n− 1)(n− 2) . . . 2 · 10! = 1

n! = n · (n− 1)!


Example

5! = 5 · 4 · 3 · 2 · 1 = 120

7!

6!=

7 · 6!6!

= 7

8!

5!=

8 · 7 · 6 · 5!5!

= 8 · 7 · 6 = 336

52!

5!47!=

52 · 51 · 50 · 49 · 48 · 47!5 · 4 · 3 · 2 · 1 · 47!

= 2, 598, 960


Example

6! = 720

10!

9!= 10

10!

7!= 720

5!

0!3!= 20

20!

3!17!= 1140


Computing principals

The sum principal.

Suppose that we want to buy a computer from one of two makesA1 and A2.

Suppose also that those makes have 12 and 18 different models,respectively.

Then how many models are there altogether to choose from ?

Since we can choose one of 12 models of make A1 or one of 18 ofA2, there are altogether 12 + 18 = 30 models to choose from.


The sum principal

Let A1 and A2 be disjoint events, that is events having no commonoutcomes, with n1 and n2 possible outcomes, respectively.

Then the total number of outcomes for the event ”A1 or A2” isn1 + n2.

Note that the events must be disjoint, that is they must nothave common outcomes for this principle to be applicable.

The general sum principal.

Let A1, A2, ...Ak be disjoint events with n1, n2, ...nk possibleoutcomes, respectively. Then the total number of outcomes for theevent ”A1 or A2 or ... or Ak” is n1 + n2 + ...+ nk.


The product principal.

A computer manufacturer offers five different CPU clock rates,four different sizes of RAMs and three different capacities of disksamong others.

How many different choices of computer there are assuming thateverything else is the same for simplicity?

First, there are five possible clock rates. Then for each selection ofclock rate one can choose one of four RAM sizes giving 5 * 4different combinations of clock rate and RAM size. Then for eachof those 20 combinations there are three different disk capacities.Thus there are 5 * 4 * 3 = 60 different combinations ofconfigurations altogether.


The product principal

Let A1 and A2 be events with n1 and n2 possible outcomes,respectively.

Then the total number of outcomes for the sequence of thetwo events is n1 × n2.

The general product principal.

Let A1, A2, ...Ak be events with n1, n2, ...nk possible outcomes,respectively. Then the total number of outcomes for the sequenceof these k events is n1 × n2 × ...× nk.


ExerciseA committee is to be chosen from among 8 mathematicians, 10physicists, 12 biologist. If the committee is to have two members ofdifferent backgrounds, how many such committees can be chosen?

8× 10 + 8× 12 + 10× 12 = 296.


ExerciseSri Manggis Restaurant serves 8 varieties of fried rice, 7 varieties ofnoodles, 10 varieties of cold drink and 9 variety of hot drink. Howmany possible choices are there if a customer wants to:

a) take a meal?

8+7=15

b) have a drink?

10+9=19

c) take a meal or have a drink?

8+7+10+9=34

d) take a meal and have a drink?

15×19=285


e) order 2 fried rice, a noodle, 3 cold drinks and 2 hot drinks

8 · 8 · 7 · 10 · 10 · 10 · 9 · 9=36,288,000

f) order either 2 fried rice or 3 noodles and either 2 cold drinksor 2 hot drinks.

(8 · 8 + 7 · 7 · 7)× (10 · 10 + 9 · 9)=73,667


Counting rules

Sample.

Suppose you want to choose a four digit pin code for your debitcard.

Each digit can take the numbers 0, 1, . . . , 9. How many possiblechoices do you have?

Since each of the 10 digits could be chosen for each position, thereare

10× 10× 10× 10 = 10000

possible choices.


Sample

If we have n distinct objects and we select r of these with replace-ment, and order is important, then there are

nr

possible outcomes.

ExampleHow many 3 letter codes can be formed if the letters can berepeated ?

There are 26 alphabets (A to Z). Therefore

26× 26× 26 = 263 = 17576.


Permutation.

Definition

A permutation of a set of objects is an arrangement of those objectsin a order without repetitions.

The number of permutations of a set with n distinct elementsis

n! = n(n− 1)(n− 2) · · · · · 2 · · · 1.

ExampleThe number of permutations of the set {1, 2, 3} is 3! = 6. Theyare 123, 132, 213, 231, 312, 321.


ExerciseIn a Science and Mathematics society’s annual meeting, threechairs are allocated for the President, Secretary and Treasurer. Inhow many ways they can arranged themselves by seating in thechairs?

3! = 6

ExerciseTo promote teaching habit among the students, Mr Idris donate 5story books for the class. In how many ways the books can bearranged in the book rack?

5! = 120


The number of k permutations of a set of n distinct elementsis

n(n− 1) · · · · · (n− k + 1) =n!

(n− k)!= nPk.

Example

The number of 2-permutations of the set {1, 2, 3} is3!

(3− 2)!= 6.

They are 12, 21, 13, 31, 23, 32.


ExerciseHow many 3 digit numbers can be chosen from the digits 1, 2, 3, 4and 5 without repeating any of these digits.

5P3 = 60

ExerciseSeven students qualified for the final round of a debatingcompetition. Three prizes have been allocated for the winners.How many possibilities are there for the winner to grab the prizes?

7P3 = 210


ExerciseKarim borrowed 5 story books from his friend. He wanted read anyone of the book and arranged the rest in the book rack. In howmany ways Karim can arrange the books?

5P4 = 120

ExerciseEn. Ismail, a badminton coach, wanted to chose two out of his fiveplayers representing the school as first single and second single. Inhow many ways the two players can be determined from the fiveplayers?

5P2 = 20


ExerciseA small committee has to be chosen, out of 10 candidates to run acommunity project. In how many ways this committee, which isconsist by a President, Secretary, Treasure and a member can bedeveloped ?

10P4 = 5040

? Exercise ?How many 3 digit odd numbers can be chosen from the digits 1 to9, without repeating any of these digits.

8× 7× 5 = 280

or8P2 ×5 P1 = 280


The number of permutations of a set with n elements ofwhich there are k kinds where

n1 + n2 . . .+ nk = n,

isn!

n1!n2! · · ·nk!.

ExampleIn how many ways can we permute the letters of the wordMISSISSIPPI?

n = 11 n1 = 1M n2 = 4I ′s n3 = 4S′s n4 = 2P ′s

Hence, number of permutations is

11!

1!4!4!2!= 34650.


ExerciseHow many permutation can be chosen from the following letters:

a) A, A, B, B, B, C

b) ORROROO

c) MALAYSIA

d) STATISTIK

e) MATEMATIK

f) KALKULUS


ExerciseHow many permutation can be chosen from the following letters:

a) A, A, B, B, B, C=6!

2!3!= 60

b) ORROROO=7!

4!3!= 35

c) MALAYSIA=8!

3!= 6720

d) STATISTIK=9!

2!3!2!= 15120

e) MATEMATIK=9!

2!2!2!= 45360

f) KALKULUS=8!

2!2!2!= 5040


ExerciseA librarian wanted to arranged 5 similar Pure Mathematics books,4 similar Mechanics books and 3 similar Statistics books in onerow of a book rack. Find the count for the possible arrangement.

12!

5!4!3!= 27720

? Exercise ?Four bricks were taken out from a box which contains 3 red bricksand 2 grey bricks. These four bricks later were arranged from leftto right. Find the count for the possible arrangement.

4!

3!1!+

4!

2!2!= 10


Combination.

Definition

A combination of a set of objects is a subset of those objects.(order is irrelevant)

The number of distinct subsets or combinations of size k thatcan be selected (without replacement) from n distinctobjects is given by

n!

(n− k)!k!= nCk.

Example

There are 3C2 =3!

(3− 2)!2!= 3 subsets of size 2 of the set

{1, 2, 3}. They are {1, 2}, {1, 3}, {2, 3}.


ExerciseIn how many ways 4 prefect can be chosen from 10 candidates?

10C4 = 210

ExerciseA school’s badminton team consist 8 students. In how many wayscan we choose a team of two?

8C2 = 28


? Exercise ?A tennis team which consist of 4 male and 3 female was chosenfrom 7 male and 5 female. In how many ways this team can bechosen?

7C4 ×5 C3 = 350

Exercise8 dots were plotted with a condition that no three dots falls in astraight line. Find the number triangles if the vertices of thetriangles were form by the those 8 dots.

8C3 = 56


? Exercise ?Two parallel line with some dots plotted on it were sketched. Inthe first line, there are two dots plotted on it. In the second line,there are 8 dots plotted on it. Find the number of triangles thatcan be form out of these dots.

2C2 ×8 C1 +2 C1 ×8 C2 = 64

? Exercise ?5 teachers were selected from 8 male teachers and 5 femaleteachers to perform duty in a examination. In how many waysthese 5 teachers can be selected with the condition that malesexceeds the females.

8C5 ×5 C0 +8 C4 ×5 C1 +

8 C3 ×5 C2 = 966


? Exercise ?5 candidates will be chosen to fill the vacancy in a company. Thereare 6 females and 5 males. In how many ways these candidates canbe chosen if it compose of

a) 3 female and 2 male6C3 ×5 C2 = 200

b) 5 peoples, regardless the gender11C5 = 462

c) At least 4 female6C4 ×5 C1 +

6 C5 ×5 C0 = 81


? Exercise ?In how many ways 10 people can be divided into

a) 2 group which consist of 7 and 3 people10C7 ×3 C3 = 120

b) 3 group which consist of 4,3 and 2 people with one of thembeen isolated.10C4 ×6 C3 ×3 C2 = 12600


? Exercise ?In how many ways 5 kids can be divided into two groups whichconsist of 3 and 2 people respectively.

5C3 ×2 C2 = 10

? Exercise ?In how many ways 9 kids can be divided into three groups whichconsist of 4, 3 and 2 people respectively.

9C4 ×5 C3 ×2 C2 = 1260


Selection.

Definition

A selection is a combination with repetition.

The number of ways of making k selections from n objectswhen selection is made with replacement and order is notimportant is

(n+ k − 1)!

(n− 1)!k!= n+k−1Ck.

ExampleIf 3 candies to be chosen among 5 kind of candies, how manydifferent selection can be made if repetition is allowed.

5+3−1C3 = 35 possible selections.


ExerciseJohn has got 1 dollar, with which he buy green, red and yellowcandies. Each candy costs 50 cents. John will spend all the moneyhe has on candies. How many different combinations of green, redand yellow candies can he buy

3+2−1C2 = 6

ExerciseSuppose in a lottery, a player selects 6 numbers from 1 to 48.During the draw, 6 numbers are drawn at random and withreplacement. To win the first prize, all 6 numbers must matchthose drawn in any order. How many winning numbers arepossible?

48+6−1C6 = 22957480


bpa 12203 mathematics for management chapter 3

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