1. Permutations and Combinations
Exercises 5.3: 4, 6, 12, 16, 30, 40
Exercises 5.5: 4, 10

2. Permutation A permutation is an ordering of objects
For example, 3 blocks can be ordered 6 ways
There are n! permutations of n elements
Easily proved using the Product rule

3. r-Permutation
r-permutations are used to order subsets of distinct elements
P(n,r) represents the r-permutations of a set of n distinct elements
P(n,r) = n(n-1)(n-2) … (n-r+1) for r  n
P(n,r) = n! / (n-r)!
Example
There are 8 different candy to choose from. How many different “meals” can you make by choosing 3 of them?
P(n,r) = P(8,3) = 8*7*6

4. r-Combination
What if it does not matter which order you eat the candy?...
A combination is an unordered arrangement of elements in a set
Example: a 3-combination from a set of 12 colored blocks

5. r-Combination
An r-combination is a selection of r unordered elements from a set (or simply a subset with r elements)
C(n,r) denotes the number of r-combinations of a set of n elements
C(n,r)= P(n,r)/r! = n!/[(n-r)!r!] = C(n,n-r)

6. Combinations with Repetition
Example
How many ways are there to select any 4 pieces of fruit from a bowl containing at least 4 apples, 4 bananas and 4 pears assuming selection order does not matter?
Solution
List the 15 possibilities

7. Combinations with Repetition
Theorem:
There are C( n + r - 1, r) r-combinations from a set with n unique elements when repetition of elements is allowed.
Previous Example:
n = 3 different fruits
r = 4 items to select
C( 3+4-1, 4) = C(6,4) = 6!/[(6-4)!4!] = 15

8. Combinations with Repetition
Example
How many ways are there to select 3 bills from a cash box containing $1, $5, $10, $20 and $50 bills?
Solution $1 $5
$10
$20
$50
n = 5
r = 3
C( 3+5-1, 3)
C( 7, 3)
* * *
* * *
** *
***

9. Combination with Repetition
Example
How many ways can 6 balls be distributed into 9 different bins?
Solution
n = 9 unique bins
r = 6 balls
C( 9+6-1, r ) = C(14, 6)

10. Permutations with Repetition
Counting permutations with repetition relies on the multiplication rule
Example:
How many strings of length r can be formed from the English alphabet?
Theorem:
The number of r-permutations of a set of n objects with repetition is nr
abc
abc
abc …
abc
= 26*26*26 … 26
= 26r r r

11. Example: Which formula to use???
How many different strings can be made by reordering the letters of the string SUCCESS?
Solution:
3 S’s, 2 C’s, 1 E and 1 U
C(7,3) to place to S’s
C(4,2) to place the C’s
C(2,1) to place the E
C(1,1) to place the U
C(7,3)C(4,2)C(2,1)C(1,1) = 7! * 4! * 2! * 1!
3!4! 2!2! 1!1! 0!0!

12. r-Permutations with Repetition
Theorem
The number of different permutations of n objects, where there are n1 identical objects of type 1, n2 identical objects of type 2 … and nt identical objects of type t, is
n! n1! n2! …nt!

13. Permutations & Combinations
For r elements selected from n distinct elements:
r-permutations P(n,r)
r-combinations C(n,r) n!
(n-r)! n!
(n-r)!r!
r-permutations with repetition nr
For n elements with n1, n2 … nt identical objects
For n unique elements in X with r selections from X
permutations with repetition of identical objects n!
n1!n2! …nt!
combinations with repetition
C(n+r-1,r)