permutations slides

7/30/2019 Permutations Slides

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Permutations

A permutation of a set ofdistinctobjects is anorderedarrangement of the objects.

Example: Permutations of{a, b, c}

How many permutations are there?

3 ways to select x {a, b, c} ofxyz.

2 ways to select y

{a, b, c} - {x} ofxyz. 1 way to select z {a, b, c} - {x, y} ofxyz.

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Example: Permutations of {a, b, c}

ab

c

b bc c

c c

a a

a ab b

abc acb bac bca cab cba

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Permutations

Theorem 1.For any integer n with n 0, the number of permutationsof a set A with n elements is n!.

Proof Sketch.

Each permutation ofA is a sequence of n elements

There are n ways of selecting the first element of the sequence.

There are n 1 ways of selecting the second element of the sequence.

There are n k ways of selecting the (k+1)-th element of the sequence.

...

Finally, there is 1 way to choose the n-th element.

By the Product Rule, there aren(n 1)(n 2) 1 = n!

ways of selecting the n elements of the sequence

That is, there are n! permutations.

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r-Permutations

An r-permutationof a set ofn elements is an

ordered selection ofr elements from the set ofn

elements.

The number ofr-permutations of a set ofn

elements is denoted

P(n, r) .

Note: Recall that the members of a set are distinct.

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Theorem 2. Ifn and r are integers and 0 r n, thenthe number ofr-permutations of a set ofn elements isgiven by

.

r-Permutations

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Proof Sketch.

An r-permutation is a sequence ofr objects.

There are n ways of selecting the first object.

There are n 1 ways of selecting the second object, regardless of how the first

object was selected. There are n k ways of selecting the (k +1)-th object, regardless of how the

previous k objects was selected.

Finally, there are n r + 1 ways of selecting the last r-th object, regardless of howthe previous r 1 objects was selected.

By the Product Rule, the total number of ways is

r-Permutations

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Corollary 1. Ifn and r are integers and 0 r n, then

Proof Sketch.

It is easy to see that

r-Permutations

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Example: r-Permutation

There are 3!/1! = 6 ways of arranging 2 members

of{a, b, c}

ab

c

b bc ca a

ab ac ba bc ca cb

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There are

5 4 3 = 60

ways of arranging 3 of the 5 letters ofBYTES.

There are

4 3 = 12

ways of arranging 3 of the 5 letters ofBYTES such that the first

letter must be B.

Explanation:

How many ways to form the sequence Bab where a, b{ Y,T,E,S }.

Example: r-Permutation

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Prove that for all integers n 2,

Proof. Suppose n is an integer that is greater than or equal to 2. ByCorollary 1,

So,

Proving Property of P(n, r)

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r-Permutations with Repetition

Theorem 3. The number ofr-permutations of a set of

n elements with repetition allowed is

nr

Proof.

There are n ways to select an element of the set for each of the r

positions in r-permutation when repetition is allowed.

By the Product Rule, there are nr r-permutations.

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Example

There are 23=8 3-permutations of the set {0,1}:

0 1

1 10

1

000

0

0

001

1

010

0

011

1

100

0

101

1

110

0

111

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