This document defines permutations and combinations and provides methods for counting them. Permutations are ordered arrangements of objects from a set, while combinations are unordered selections. The number of r-permutations of a set with n elements is P(n,r) and can be calculated using factorials. The number of r-combinations is C(n,r) and can also be computed factorially. Several properties of permutations and combinations are proved, including Pascal's identity and Vandermonde's identity.

1. Permutations and Combinations
Rosen 4.3

2. 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…

3. 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)
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4. 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).

5. 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)!

6. 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)
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7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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)!

14. Binomial Coefficient
Another notation for C(n,r) is . 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.

15. 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:
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16. Let n be a positive integer. Then
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.
counts the number of subsets
of every size from 0 (empty
set) to n. Therefore the sum must add up to
2n.

17. Vandermonde’s Identity
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.

18. 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

19. Pascal’s Triangle
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n’th row, Cnk =k = 0, 1, …, n

20. Binomial Theorem
Let x and y be variables and let n be a positive
integer. Then