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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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