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

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

Permutation packet page 1 and 2 Homework Permutation packet page 1 and 2

Permutations

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

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

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 =

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.

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.

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.

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.

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 !

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

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?

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

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

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