MATH 105: Finite Mathematics 6-4:...

Types of Permutations Factorials Applications of Permutations Conclusion

MATH 105: Finite Mathematics6-4: Permutations

Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006

Outline

1 Types of Permutations

2 Factorials

3 Applications of Permutations

4 Conclusion

Outline

2 Factorials

4 Conclusion

Permutations

In this section we will discuss a special counting technique which isbased on the multiplication principle. This tool is called apermutation.

Permutations

A permutation is an ordered arrangement of r objects chosen fromn available objects.

Objects may be chosen with, or without, replacement. In eithercase, permutations are really special cases of the multiplicationprinciple.

Permutations

Permutations with Replacement

Example

U.S. zip codes consist of an ordering of five digits chosen from 0-9with replacement (i.e. numbers may be reused). How many zipcodes are in the set Z of all possible zip codes?

Formula

The number of ways to arrange r items chosen from n withreplacement is:

Example

c(Z ) = permutation of 5 digits from 10 with replacement

= 10 · 10 · 10 · 10 · 10

= 100, 000

Formula

Example

c(Z ) = permutation of 5 digits from 10 with replacement

= 10 · 10 · 10 · 10 · 10

= 100, 000

Formula

Permutations without Replacement

Example

If there are ten plays ready to show, and 6 time slots available, if Sis the set of all possible play schedules, what is c(S)?

Formula

The number of ways to arrange r items chosen from n withoutreplacement is:

P(n, r) =n!

(n − r)!

Example

c(S) = permutation of 6 plays from 10 without replacement

= 10 · 9 · 8 · 7 · 6 · 5= 151, 200

Formula

P(n, r) =n!

(n − r)!

Example

c(S) = permutation of 6 plays from 10 without replacement

= 10 · 9 · 8 · 7 · 6 · 5= 151, 200

Formula

P(n, r) =n!

(n − r)!

Outline

2 Factorials

4 Conclusion

Definition of a Factorial

The formula n! stands for “n factorial”. This will be useful inmany counting formulas.

Factorial

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

Example

0! = 1

Factorial

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

Example

0! = 1

Factorial

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

Example

0! = 1

Factorial

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

Example

0! = 1

1! = 1

Factorial

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

Example

0! = 1

1! = 1

2! = 2 · 1 = 2

Factorial

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

Example

0! = 1

1! = 1

2! = 2 · 1 = 2

3! = 3 · 2 · 1 = 3 · 2! = 6

Factorial

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

Example

0! = 1 4! = 4 · 3! = 24

1! = 1

2! = 2 · 1 = 2

3! = 3 · 2 · 1 = 3 · 2! = 6

Factorial

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

Example

0! = 1 4! = 4 · 3! = 24

1! = 1 5! = 5 · 4! = 120

2! = 2 · 1 = 2

3! = 3 · 2 · 1 = 3 · 2! = 6

Factorial

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

Example

0! = 1 4! = 4 · 3! = 24

1! = 1 5! = 5 · 4! = 120

2! = 2 · 1 = 2 . . .

3! = 3 · 2 · 1 = 3 · 2! = 6

Factorial

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

Example

0! = 1 4! = 4 · 3! = 24

1! = 1 5! = 5 · 4! = 120

2! = 2 · 1 = 2 . . .

3! = 3 · 2 · 1 = 3 · 2! = 6 n! = n(n − 1)!

Computing Permutations using Factorials

We can use factorials to compute the number of ways to schedulethe plays mentioned in a previous example.

Example

This may seem more complicated than necessary, but it issometimes useful to have a formula with which to work.

Example

P(10, 6) =10!

(10 − 6)!=

10 · 9 · 8 · 7 · 6 · 5 ·��4!

��4!

= 10 · 9 · 8 · 7 · 6 · 5= 151, 200

Example

P(10, 6) =10!

(10 − 6)!=

10 · 9 · 8 · 7 · 6 · 5 ·��4!

��4!

= 10 · 9 · 8 · 7 · 6 · 5= 151, 200

More Permutation Computations

Example

Use factorials to compute each permutation.

1 Find P(7, 3)

2 Find P(9, 1)

3 Find P(4, 4)

Example

1 Find P(7, 3)

2 Find P(9, 1)

3 Find P(4, 4)

Example

1 Find P(7, 3)

P(7, 3) =7!

(7 − 3)!=

7 · 6 · 5 · 4!

4!= 7 · 6 · 5 = 210

2 Find P(9, 1)

3 Find P(4, 4)

Example

1 Find P(7, 3)

P(7, 3) =7!

(7 − 3)!=

7 · 6 · 5 · 4!

4!= 7 · 6 · 5 = 210

2 Find P(9, 1)

3 Find P(4, 4)

Example

1 Find P(7, 3)

P(7, 3) =7!

(7 − 3)!=

7 · 6 · 5 · 4!

4!= 7 · 6 · 5 = 210

2 Find P(9, 1)

P(9, 1) =9!

(9 − 1)!=

9 · 8!

3 Find P(4, 4)

Example

1 Find P(7, 3)

P(7, 3) =7!

(7 − 3)!=

7 · 6 · 5 · 4!

4!= 7 · 6 · 5 = 210

2 Find P(9, 1)

P(9, 1) =9!

(9 − 1)!=

9 · 8!

3 Find P(4, 4)

Example

1 Find P(7, 3)

P(7, 3) =7!

(7 − 3)!=

7 · 6 · 5 · 4!

4!= 7 · 6 · 5 = 210

2 Find P(9, 1)

P(9, 1) =9!

(9 − 1)!=

9 · 8!

3 Find P(4, 4)

P(4, 4) =4!

(4 − 4)!=

1= 4 · 3 · 2 · 1 = 24

General Permutation Rules

The last two examples of the previous slide are examples of generalrules for permutations.

Permutation Rule #1

For any n,P(n, 1) = n

Permutation Rule #2

For any n,P(n, n) = n!

Permutation Rule #1

Permutation Rule #2

Permutation Rule #1

Permutation Rule #2

Outline

2 Factorials

4 Conclusion

Codes and Committees

Example

How many different 4-letter codes can be formed using the lettersA,B,C,D,E, and F with no repetition?

Example

A committee of 7 people wisht to select a subcommittee of 3,including a chairman and secretary for the subcommittee. In howmany ways can this be done?

Example

P(6, 4) = 6 · 5 · 4 · 3 = 360

Example

P(6, 4) = 6 · 5 · 4 · 3 = 360

Example

P(6, 4) = 6 · 5 · 4 · 3 = 360

Example

P(7, 3) = 7 · 6 · 5 = 210

More Committees

Example

A play involving 2 male and 1 female parts is to be cast from apool of 6 male and 4 female actors. How many casts are possible?

More Committees

Example

Male: P(6, 2) = 6 · 5 = 30

More Committees

Example

Male: P(6, 2) = 6 · 5 = 30

Female: P(4, 1) = 4

More Committees

Example

Male: P(6, 2) = 6 · 5 = 30

Female: P(4, 1) = 4

Combined: P(6, 2) · P(4, 1) = 30 · 4 = 120

Arranging Letters in a Word

One application of permutations is rearranging letters in a word.

Example

How many new “words” can be formed from the letters in theword “Monday”?

Be Careful!

What about the word “fell”?

Example

Be Careful!

Example

P(6, 6) = 6! = 720

Be Careful!

Example

P(6, 6) = 6! = 720

Be Careful!

Words with Repeated Letters

Example

In how many ways can the letters in the word “fell” be arranged?

Example

In how many ways can the letters in the words “ninny” and“Mississippi” be arranged?

Example

P(4, 4)

P(2, 2)=

2!= 12

Example

P(4, 4)

P(2, 2)=

2!= 12

Example

P(4, 4)

P(2, 2)=

2!= 12

Example

P(5, 5)

P(3, 3)=

3!= 20

Example

P(4, 4)

P(2, 2)=

2!= 12

Example

P(11, 11)

P(4, 4) · P(4, 4) · P(2, 2)=

4! · 4! · 2!= 34650

Outline

2 Factorials

4 Conclusion

Important Concepts

Things to Remember from Section 6-4

1 Order matters in a permutation

2 Formulas: nr with replacement and n!(n−r)! without.

3 Arranging letters in words: watch out for repetitions!

Important Concepts

Next Time. . .

Permutations are the first of two counting rules which build on themultiplication principle. In the next section, we will introduce“combinations” in which we care only about the objects selected,and not the order in which the selection is made.

For next time

Read Section 6-5 (pp 343-349)

Do Problem Sets 6-4 A,B

Next Time. . .

Permutations are the first of two counting rules which build on themultiplication principle. In the next section, we will introduce“combinations” in which we care only about the objects selected,and not the order in which the selection is made.

For next time

Read Section 6-5 (pp 343-349)

Do Problem Sets 6-4 A,B

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