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CHAPTER 1

Combinatorial Analysis

Written by:Samah sindi

& Shafiaa alhodairah

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A) Basic Counting Principle

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A) Basic Counting Principle

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A) The basic principle of counting

Ex. 2 : A small community consists of 10 women

, each of women has 3 children , if one women

and one of the children are to be chosen as

mother and child of the year , how many

different choices are possible ?

Solution:

m=10, n=30

choices=10*30=300.

A) The basic principle of counting

Ex. 2 : A small community consists of 10 women ,

each of women has 3 children , if one women

and one of her children are to be chosen as

mother and child of the year , how many

different choices are possible ?

Solution:

m=10, n=3

choices=10*3=30.

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A) Basic Counting Principle

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A) Basic Counting Principle

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A) Basic Counting Principle

Ex. 4 :A college planning committee consists

of 3 fresh man. 4 sophomores, 5 juniors, and 2

seniors, a sub committee of 4, consisting a

single representative from each of the classes

is to be chosen. How many different

subcommittees are possible?

Solution:

120.2543Total

2.n5,n4,n3,n 4321

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B) Permutations

Rule1:

A permutation of n different objects is an arrangement

of these n objects.

The number of permutations of n different objects

taken all at a time is n!

The number of permutations of r different objects

taken from n objects where ( 0 ≤ r ≤ n ) is noted by:

!

!

rn

nnP r

n

r

B) Permutations

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B) Permutations

Ex. 5 :How many different arrangements are

possible to arrange 10 people?

Solution:

Ex. 6 : How many different batting orders are

possible for a baseball team consisting of 9

players?

Solution:

.3628800123456789210!10

.362880123456789!9

B) Permutations

Ex. 7: From group of 6 Americans, 5 French and 4

Russians is to be ordered in row. Determine the

number of possible orderings if no restrictions are

imposed.

Number of orderings= 15!

Ex. 8: From group of 6 Americans, 5 French and 4

Russians is to be ordered in row. Determine the

number of possible Americans-French-Russians

orderings, if each nationality is to be ordered

within itself.

Number of orderings= 6!x 5! x 4!=2073600

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B) Permutations

Ex. 9: From group of 6 Americans, 5 French

and 4 Russians is to be ordered in row.

Determine the number of possible orderings

in any order, if each nationality is to be

ordered within itself.

Number of orderings= 6!x 5! x 4! x 3!

= 12441600

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B) Permutations

!n ... !n !n

n!

n ... n n n

n

k21k321

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B) Permutations

Ex. 10 : How many different letter arrangements

can be formed using the letters P E P P E R?

Solution:

t.arrangemenletterpossible602!3!

6!

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arethereThen,

6.123nnnn

1.Rn

2,En

3,Pn

321

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B) Permutations

Ex. 11 : How many different signals, each

consisting of 9 flags hung in a line, can be made

from a set of 4 white flags,3 red flags, and 2 blue

flags if all flags of the same color are identical?

Solution:

signals.different12602!3!4!

9!

234

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arethereThen,

9.234nnnn

2.flagsbluen

3,flagsredn

4,flagswhiten

321

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2

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C) Combinations

The number of different subgroups of size r

that can be chosen from a set of size n is

given by:

nr

r

P

rnr

n

r

nC

n

rn

r

,

!!!

!

C) Combinations

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C) Combinations

Ex. 12 :A committee of 3 people is to be

performed from a group of 20 people.

How many different committees are

possible?

Solution:

.committeespossible114017!3!

20!

3

20

areThere

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C) Combinations

Ex. 13 :From a group of 5 women and 7

men, How many different committees

consisting of 2 women and 3 men can be

performed ?

Solution:

.committeespossible3504!3!

7!

3!2!

5!CC

areThere

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3

5

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C) Combinations

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C) Combinations

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