PowerPoint Template Title: PowerPoint Template Author: jj Created Date: 2/20/2017 12:49:01 PM
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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.
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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 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
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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
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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!
23
6
arethereThen,
6.123nnnn
1.Rn
2,En
3,Pn
321
3
2
1
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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
9
arethereThen,
9.234nnnn
2.flagsbluen
3,flagsredn
4,flagswhiten
321
3
2
1
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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
,
!!!
!
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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
7
3
5
2
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C) Combinations
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C) Combinations
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