1. Warm-Up #4 Monday, 2/8/2016 Simplify 19! 13!

2. Warm-Up #5 , Tuesday 2/9/2016 Simplify 12! 9!

3. Permutation packet page 1 and 2
Homework
Permutation packet page 1 and 2

4. Permutations

5. The product can be written as a factorial.
A permutation is an arrangement of things in a certain order. (THE ORDER DOES MATTER!)
If no letter can be used more than once, there are 6 permutations of the first 3 letters of the alphabet: ABC, ACB, BAC, BCA, CAB, and CBA.
first letter ?
second letter ?
third letter ?
3 choices
2 choices
1 choice • •
The product can be written as a factorial.
3 • 2 • 1 = 3! = 6

6. Course 3
By definition, 0! = 1.
Remember!

7. Notice that the product can be written as a quotient of factorials.
Course 3
If no letter can be used more than once, there are 60 permutations of the first 5 letters of the alphabet, when taken 3 at a time: ABE, ACD, ACE, ADB, ADC, ADE, and so on.
first letter ?
second letter ?
third letter ?
5 choices
4 choices
3 choices  
= 60 permutations
Notice that the product can be written as a quotient of factorials.
5 • 4 • 3 • 2 • 1
2 • 1 =
5! 2!
60 = 5 • 4 • 3 =

8. Additional Example 2A: Finding Permutations
Course 3
Additional Example 2A: Finding Permutations
Jim has 6 different books.
Find the number of orders in which the 6 books can be arranged on a shelf.
The number of books is 6. 6!
(6 – 6)! = 6! 0! =
6 • 5 • 4 • 3 • 2 • 1 1 =
6P6 =
720
The books are arranged 6 at a time.
There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

9. Additional Example 2B: Finding Permutations
Course 3
Additional Example 2B: Finding Permutations
If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged.
The number of books is 6. 6!
(6 – 3)! = 6! 3! =
6 • 5 • 4 • 3 • 2 • 1
3 • 2 • 1 =
6P3 =
6 • 5 • 4
The books are arranged 3 at a time.
= 120
There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

10. Permutations and Combinations
Course 3
Permutations and Combinations
Check It Out: Example 2A
There are 7 soup cans in the pantry.
Find the number of orders in which all 7 soup cans can be arranged on a shelf.
The number of cans is 7. 7!
(7 – 7)! = 7! 0! =
7 • 6 • 5 • 4 • 3 • 2 • 1 1
7P7 =
= 5040
The cans are arranged 7 at a time.
There are 5040 orders in which to arrange 7 soup cans.

11. Permutations and Combinations
Course 3
Permutations and Combinations
Check It Out: Example 2B
There are 7 soup cans in the pantry.
If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.
The number of cans is 7. 7!
(7 – 4)! = 7! 3! =
7 • 6 • 5 • 4 • 3 • 2 • 1
3 • 2 • 1
7P4 =
The cans are arranged 4 at a time.
= 7 • 6 • 5 • 4
= 840
There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

12. Permutations with Identical Objects
The number of distinct permutations of n objects with r identical objects is given by
𝒏! 𝒓! , where 1≤𝑟≤𝑛
More identical objects:
𝒏! 𝒓 𝟏 ! 𝒓 𝟐 ! 𝒓 𝟑 !

13. Example 1
Shay is planting 11 colored flowers in a line. In how many ways can she plant 4 red flowers, 5 yellow flowers, and 2 purple flowers? 11! 4!∙5!∙2! = Answer: 6930 ways

14. Example 2
Kelly is planting 10 colored flowers in a line. In how many ways can she plant 1, green, 1 yellow, 1 orange, 1 brown, 4 red flowers and 3 purple flowers?

15. Circular Permutations
If n distinct objects are arranged around a circle, then there are (n-1)! Circular permutations of the n objects.

16. Example 3
In how many different ways can 7 different appetizers be arranged on a circular tray? Answer: 720 distinct ways

17. Example 4
Five different stuffed animals are to be placed on a circular display rack in a department store. In how many ways can this be done?
Answer: 24 ways