1. 6-7 Permutations & Combinations M11.E.3.2.1: Determine the number of permutations and/or combinations or apply the fundamental counting principle

2. Objectives Permutations Combinations

3. Vocabulary A permutation is an arrangement of items in a particular order. n factorial For any positive integer n, n! = n(n – 1) · … · 3 · 2 · 1 For n = 0, n! = 1

4. Vocabulary Fundamental Counting Principle If there are a ways the first event can occur and b ways the second event can occur, then there are a x b ways that both events can occur. Example: You are ordering dinner at a restaurant. You have a choice of soup or salad for an appetizer. A choice of steak, chicken, or tofu for a main entrée, and a choice of pie or ice cream for dessert. How many different meals can you have?

5. In how many ways can 6 people line up from left to right for a group photo? Since everybody will be in the picture, you are using all the items from the original set. You can use the Multiplication Counting Principle or factorial notation. There are six ways to select the first person in line, five ways to select the next person, and so on. The total number of permutations is 6 5 4 3 2 1 = 6!. 6! = 720 The 6 people can line up in 720 different orders. Finding Permutations

6. Vocabulary Permutations If order does matter, then you are working with permutations. The number of permutations of n items of a set arranged r items at the time is n P r Ex. Seven people are running a race. How many different outcomes of first, second, and third place are possible? n = 7 (total number of items) and r = 3 (how many we’re choosing) This is written 7 P 3

7. Permutations n P r = for 0 ≤ r ≤ n 10 P 5 = = = 5040

8. How many 4-letter codes can be made if no letter can be used twice? Method 1:Use the Multiplication Counting Principle. 26 25 24 23 = 358,800 Method 2:Use the permutation formula. Since there are 26 letters arranged 4 at a time, n = 26 and r = 4. There are 358,800 possible arrangements of 4-letter codes with no duplicates. 26 P 4 = = = 358,800 26! (26 – 4)! 26! 22! Real World Example

9. Vocabulary Combinations If order does not matter, then you are working with permutations. The number of combination of n items of a set arranged r items at the time is n C r Ex. Seven people are running a race. How many different ways can 3 people receive a medal? (It doesn’t matter whether they get a gold or bronze) n = 7 (total number of items) and r = 3 (how many we’re choosing) This is written 7 C 3

10. Combinations n C r = for 0 ≤ r ≤ n 5 C 3 = = = = 10

11. Evaluate 10 C 4. 10 9 8 7 4 3 2 1 = = 210 10 C 4 = 10! 4!(10 – 4)! = 10! 4! 6! 10 9 8 7 4 3 2 1 = 654321 654321 Finding Combinations

12. A disk jockey wants to select 5 songs from a new CD that contains 12 songs. How many 5-song selections are possible? Relate: 12 songs chosen 5 songs at a time Define: Let n = total number of songs. Let r = number of songs chosen at a time. You can choose five songs in 792 ways. Use the n C r feature of your calculator. Write: n C r = 12 C 5 Real World Example

13. A pizza menu allows you to select 4 toppings at no extra charge from a list of 9 possible toppings. In how many ways can you select 4 or fewer toppings? The total number of ways to pick the toppings is 126 + 84 + 36 + 9 + 1 = 256. There are 256 ways to order your pizza. You may choose 4 toppings, 3 toppings, 2 toppings, 1 toppings, or none. 9 C 49 C 39 C 29 C 19 C 0 Real Worl Example