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Click on the picture. Main Menu (Click on the topics below). Combinations Example Example Example Example Example Example Theorem. Combinations. Sanjay Jain, Lecturer, School of Computing. Let n, r 0, be such that r n. - PowerPoint PPT Presentation
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Main Menu Main Menu (Click on the topics below)
Combinations
Example
Example
Example
Example
Example
ExampleTheorem
Click on the picture
CombinationsCombinations
Sanjay Jain, Lecturer, Sanjay Jain, Lecturer,
School of ComputingSchool of Computing
CombinationsCombinations
Let n, r 0, be such that r n. Suppose A is a set of n elements. An r-combination
of A, is a subset of A of size r.
r
n Pronounced: n choose rDenotes the number of different r-combinations of a set of size n.
Some other notations commonly used are nCr, and C(n,r).
CombinationsCombinations
Combinations ---> unordered selectionPermutation ---> ordered selection
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ExampleExample
There are 7 questions in an exam.You need to select 5 questions to answer.How many ways can you select the questions to
answer?(order does not matter)
5
7 Here 7 is the number of questions and 5 is the number of questions selected
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ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T1: give 13 cards to N
T2: give 13 of the remaining cards to E
T3: give 13 of the remaining cards to S
T4: give 13 of the remaining cards to W
ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T1: give 13 cards to N
T1: can be done in ways.
13
52
ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T2: give 13 of the remaining cards to E
T2: can be done in ways.
13
39
ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T3: give 13 of the remaining cards to S
T3: can be done in ways.
13
26
ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T4: give 13 of the remaining cards to W
T4: can be done in ways.
13
13
ExampleExample
In how many ways 52 cards can be distributed to N, S, E, W in a game of bridge.
T1: give 13 cards to N
T2: give 13 of the remaining cards to E
T3: give 13 of the remaining cards to S
T4: give 13 of the remaining cards to W
Thus using the multiplication rule total number of ways in which cards can be distributed is
13
13*
13
26*
13
39*
13
52
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ExampleExample
70 faculty members.Need to choose two committees:
A) Curriculum committee of size 4
B) Exam committee of size 3
How many ways can this be done if the committees are to be disjoint?
ExampleExampleT1: Choose curriculum committee
T2: Choose exam committee
T1:
4
70 ways
T2:
3
66ways
The selection of both comm can be done in
4
70
3
66* ways
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ExampleExample
From 300 students I need to select a president, secretary and 3 ordinary members of Executive committee.
How many ways can this be done?
1st Method:
T1: president---300
T2: secretary---299
T3: 3 ordinary members---298C3
3
298*299*300
2nd Method:
T1: 5 members of the committee --- 300C5
T2: choose president among the members of the committee --- 5
T3: choose secretary among the members of the committee --- 4
4*5*5
300
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ExampleExample
From 300 students I need to select a football team of 11 players.
Tom and Sam refuse to be in the team together.How many ways can the team be selected?
Case 1: Tom is in the team.Case 2: Sam is in the team.Case 3: Neither Tom nor Sam is in the team.
ExampleExample
Case 1: Tom is in the team.
10
298
Need to select 10 out the remaining 298 students.
ExampleExample
Case 2: Sam is in the team.
10
298
Need to select 10 out the remaining 298 students.
ExampleExample
Case 3: Both Tom and Sam are not in the team.
11
298
Need to select 11 out the remaining 298 students.
ExampleExample
From 300 students I need to select a football team of 11 players.
Tom and Sam refuse to be in the team together.How many ways can the team be selected?
Case 1: Tom is in the team. --- 298C10
Case 2: Sam is in the team. --- 298C10
Case 3: Neither Tom nor Sam is in the team. --- 298C11
11
298
10
298
10
298
Thus total number of possible ways to select the team is:
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ExampleExample
There are 6 boys and 5 girls. In how many ways can one form an executive committee of size 4 such that there is at least one member of each sex?
Wrong method:
T1: select one boy. --- 6 ways
T2: select one girl. --- 5 ways
T3: select 2 others. --- 9C2 ways
6*5* 9C2 ways to select the committee.
ExampleExample
There are 6 boys and 5 girls. In how many ways can one form an executive committee of size 4 such that there is at least one member of each sex?
Wrong method:
B1G1G2, G3
B1G2G1, G3
B1G3G1, G2
Selection of B1, G1, G2, G3 is counted as:
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ExampleExample
There are 6 boys and 5 girls. In how many ways can one form an executive committee of size 4 such that there is at least one member of each sex?
ExampleExampleCorrect method: A: Choose 4 members of the committee (without
restrictions) B: Choose 4 members of the committee without any
boys. C: Choose 4 members of the committee without any
girls.D: Choose 4 members of the committee with at least
one boy and at least one girl.
D=A-B-C
4
11
4
5
4
6
4
6
4
5
4
11
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TheoremTheorem
Proof:
)!(!
!
!
),(
knk
n
k
knP
k
n
Choose k out of n elements
Choose k out of n elements in order
a) Choose k out of n elements.b) Put order
P(n,k)
k
n
Thus:
!*),( kk
nknP
k
n
!k
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