1. Permutations and Combinations

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)

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)

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. 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?
exactly 2 vowels
at least 1 vowel
at least 2 vowels

9. The English alphabet contains 21 consonants and 5 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

10. The English alphabet contains 21 consonants and 5 vowels
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
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.

12. The English alphabet contains 21 consonants and 5 vowels
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.
*5*215

13. Corollary 1: Let n and r be nonnegative integers with r  n
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:

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 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
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