# Permutations and Combinations

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```					Permutations and Combinations

Rosen 4.3
Permutations
• A permutation of a set of distinct objects is
an ordered arrangement these objects.
• An ordered arrangement of r elements of a
set is called an r-permutation.
• The number of r-permutations of a set with
n elements is denoted by P(n,r).
A = {1,2,3,4} 2-permutations of A include
1,2; 2,1; 1,3; 2,3; etc…
Counting Permutations
• Using the product rule we can find P(n,r)
= n*(n-1)*(n-2)* …*(n-r+1)
= n!/(n-r)!
How many 2-permutations are there for the
set {1,2,3,4}? P(4,2)
4 * 3 * 2 *1 4!
4*3                12
2 *1     2!
Combinations
• An r-combination of elements of a set is an
unordered selection of r element from the set.
(i.e., an r-combination is simply a subset of the set
with r elements).
Let A={1,2,3,4} 3-combinations of A are
{1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}(same as {3,2,4})
• The number of r-combinations of a set with n
distinct elements is denoted by C(n,r).
Example
Let A = {1,2,3}
2-permutations of A are: 1,2 2,1 1,3 3,1 2,3 3,2
6 total. Order is important

2-combinations of A are: {1,2}, {1,3}, {2,3}
3 total. Order is not important
If we counted the number of permutations of each 2-
combination we could figure out P(3,2)!
How to compute C(n,r)
• To find P(n,r), we could first find C(n,r),
then order each subset of r elements to
count the number of different orderings.
P(n,r) = C(n,r)P(r,r).
• So C(n,r) = P(n,r) / P(r,r)
n!
(n  r )! n!(r  r )!         n!
                           
r!             (n  r )! r! r!(n  r )!
(r  r )!
A club has 25 members.
• How many ways are there to choose four members
of the club to serve on an executive committee?
– Order not important
– C(25,4) = 25!/21!4! = 25*24*23*22/4*3*2*1
=25*23*22 = 12,650
• How many ways are there to choose a president,
vice president, secretary, and treasurer of the club?
– Order is important
– P(25,4) = 25!/21! = 303,600
The English alphabet contains 21 consonants and
5 vowels. How many strings of six lower case
letters of the English alphabet contain:
•   exactly one vowel?
•   exactly 2 vowels
•   at least 1 vowel
•   at least 2 vowels
The English alphabet contains 21 consonants and
5 vowels. How many strings of six lower case
letters of the English alphabet contain:
• exactly one vowel?
Note that strings can have repeated letters!
We need to choose the position for the vowel
C(6,1) = 6!/1!5! This can be done 6 ways.
Choose which vowel to use.
This can be done in 5 ways.
Each of the other 5 positions can contain any of the 21
consonants (not distinct).
There are 215 ways to fill the rest of the string.
6*5*215
The English alphabet contains 21 consonants and
5 vowels. How many strings of six lower case
letters of the English alphabet contain:
• exactly 2 vowels?
Choose position for the vowels.
C(6,2) = 6!/2!4! = 15
Choose the two vowels.
5 choices for each of 2 positions = 52
Each of the other 4 positions can contain any of 21
consonants.
214
15*52*214
The English alphabet contains 21 consonants and
5 vowels. How many strings of six lower case
letters of the English alphabet contain:
• at least 1 vowel
Count the number of strings with no vowels
and subtract this from the total number of
strings.
266 - 216
The English alphabet contains 21 consonants and
5 vowels. How many strings of six lower case
letters of the English alphabet contain:
• at least 2 vowels
Compute total number of strings and subtract
number of strings with no vowels and the
number of strings with exactly 1 vowel.
266 - 216 - 6*5*215
Corollary 1: Let n and r be nonnegative
integers with r  n. Then C(n,r) = C(n,n-r)

Proof:
C(n,r) = n!/r!(n-r)!
C(n,n-r) = n!/(n-r)!(n-(n-r))! = n!/r!(n-r)!
Binomial Coefficient
n 

Another notation for C(n,r) is r  . This
number is also called a binomial coefficient.

These numbers occur as coefficients in the
expansions of powers of binomial
expressions such as (a+b)n.
Pascal’s Identity
Let n and k be positive integers with n  k.
Then C(n+1,k) = C(n, k-1) + C(n,k).
Proof: C (n, k  1)  C (n, k )
n!               n!
                      
(k  1)!(n  k  1)! k!(n  k )!
kn!                   (n  k  1)n!
                               
k (k  1)!(n  k  1)(n  k )! k!(n  k  1)(n  k )!
n!(k  n  k  1)      n!(n  1)         (n  1)!
                                    
k!(n  1  k )!    k!(n  1  k )! k!(n  1  k )!
 C (n  1, k )
n

Let n be a positive integer. Then  C(n,k)  2
n

k 0

Proof: We know from set theory that the
number of subsets in a set of size n is 2n.
We also know that C(n,k) is the number of
subsets of a set of size n that are of size k.
n

0C(n,k) counts the number of subsets
k
of every size from 0 (empty
set) to n. Therefore the sum must add up to
2 n.
Vandermonde’s Identity
r
C(m  n, r)   C(m, r  k)C(n,k).
k 0

Proof: Suppose there are n items in one set and m items in
a second set. Then the total number of ways to pick r
elements from the union of these sets is C(m+n,r).
Another way to pick r elements from the union is to pick k
elements from the first set and then r-k elements from the
second set, where 0  k  r. There are C(n,k) ways to pick
the k elements from the first set and C(m,r-k) ways to pick
the rest of the elements from the second set.
r
C(m  n, r)   C(m, r  k)C(n,k).
k 0

Proof: Suppose there are n items in one set and m items in a
second set. Then the total number of ways to pick r
elements from the union of these sets is C(m+n,r).
Another way to pick r elements from the union is to pick k
elements from the first set and then r-k elements from the
second set, where 0  k  r. For any k,there are C(n,k)
ways to pick the k elements from the first set and C(m,r-k)
ways to pick the rest of the elements from the second set.
By the product rule there are C(m,r-k)C(n,k) ways to pick r
elements for a particular k. For all possible values of k
r

 C(m, r  k)C(n, k).
k 0
Pascal’s Triangle
0 
                                                 1
0 
1 
                1

0             1                                 1           1

2 
                 2 
                2 
                       1           2       1
0              1             2 

3             3
                 3 
                3
          1       3           3       1

0             1              2             3
1       4           6       4       1
n 
n’th row, Cnk =k 0, 1, …, n
=
r 
Binomial Theorem
Let x and y be variables and let n be a positive
integer. Then
n
(x  y)   C(n, j)x
n                nj       j
y
j 0

n n n n1 n n 2 2  n  n 1 n n
  x   x y   x y  ...         xy   y
0   1     2        n  1     n

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