Permutations and Combinations Course 3 10-9 Permutations and Combinations Warm Up #20 Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 3. How many different 4–digit phone extensions are possible? 16 12 10,000

Homework Solution. Lesson 10.1 10) 1/5 12) 1/6 14) 1/2 16) 1/3 18) 0 26) 17,576,000 27) 6,760,000 28) 17,576,000 29) 6,760,000

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Lesson 10.2 Permutations

5! = 5 • 4 • 3 • 2 • 1 10-9 Permutations and Combinations Course 3 10-9 Permutations and Combinations The factorial of a number is the product of all the whole numbers from the number down to 1. ** 0! = 1 5! = 5 • 4 • 3 • 2 • 1 Read 5! as “five factorial.” Reading Math

Additional Example 1: Evaluating Expressions Containing Factorials Course 3 10-9 Permutations and Combinations Additional Example 1: Evaluating Expressions Containing Factorials Evaluate each expression. A. 9! 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880 8! B. 6! Write out each factorial and simplify. 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1 Multiply remaining factors. 8 • 7 = 56

Additional Example 1: Evaluating Expressions Containing Factorials Course 3 10-9 Permutations and Combinations Additional Example 1: Evaluating Expressions Containing Factorials 10! (9 – 2)! C. 10! 7! Subtract within parentheses. 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7  6  5  4  3  2  1 10 • 9 • 8 = 720

Permutations and Combinations Course 3 10-9 Permutations and Combinations Check It Out: Example 1 9! (8 – 2)! C. 9! 6! Subtract within parentheses. 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6  5  4  3  2  1 9 • 8 • 7 = 504

Permutations and Combinations Course 3 10-9 Permutations and Combinations 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

Permutations and Combinations Course 3 10-9 Permutations and Combinations By definition, 0! = 1. Remember!

Permutations and Combinations Course 3 10-9 Permutations and Combinations 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 10-9 Permutations and Combinations 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 10-9 Permutations and Combinations 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 10-9 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 10-9 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 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

Review on Permutations

Homework Solution lesson 10.2 45) 151,200 46) 34,650 47) 908,107,200 48) 24 49) 720 50) 24

Homework Lesson 10.2_pg 640 #12-22 EVEN

Classwork Write down your name, period, date TITLE: Lesson 10.2 Classwork Task: your task is to complete each problem. You need to show your work in order to get full classwork points

Example 10) Kelly is planting 10 colored flowers in a line. In how many ways can she plant 4 red flowers and 3 purple flowers? 10) 10! 3!4! =25200 ways

Warm-Up #21 7P3= 4P3 8P3 = In how many ways can 3 males (Joe, Jerry John) and 3 females (Jamie, Jenny, Jasmine) be seated in a row if the genders alternate down the row? 210 0.7 72

Homework Solution lesson 10.2 12) 2 14) 210 16) 720 18) 210 20) 1000 22) About 0.07

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Lesson 10.3 Combinations

Example You have 3 letters. A B C. Write all of the combinations. A B C A C A B A C B C A C A B C B A How many different combinations can I make if the order of the letters does matter? Permutations: 6 ways How many different combinations can I make if the order of the letters does not matter? Combination: 1 way

Course 3 10-9 Permutations and Combinations A combination is a selection of things in any order. The order does not matter.

Permutations and Combinations Course 3 10-9 Permutations and Combinations

Additional Example 3A: Finding Combinations Course 3 10-9 Permutations and Combinations Additional Example 3A: Finding Combinations Mary wants to join a book club that offers a choice of 10 new books each month. If Mary wants to buy 2 books, find the number of different pairs she can buy. 10 possible books 10! 2!(10 – 2)! = 10! 2!8! 10C2 = 2 books chosen at a time 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) = = 45 There are 45 combinations. This means that Mary can buy 45 different pairs of books.

Additional Example 3B: Finding Combinations Course 3 10-9 Permutations and Combinations Additional Example 3B: Finding Combinations If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy. 10 possible books 10! 7!(10 – 7)! = 10! 7!3! 10C7 = 7 books chosen at a time 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (7 • 6 • 5 • 4 • 3 • 2 • 1)(3 • 2 • 1) = = 120 There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

Permutations and Combinations Course 3 10-9 Permutations and Combinations Check It Out: Example 3A Harry wants to join a DVD club that offers a choice of 12 new DVDs each month. If Harry wants to buy 4 DVDs, find the number of different sets he can buy. 12 possible DVDs 12! 4!(12 – 4)! = 12! 4!8! 12C4 = 4 DVDs chosen at a time = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) = 495

Check It Out: Example 3A Continued Course 3 10-9 Permutations and Combinations Check It Out: Example 3A Continued There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

Permutations and Combinations Course 3 10-9 Permutations and Combinations Check It Out: Example 3B If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy. 12 possible DVDs 12! 11!(12 – 11)! = 12! 11!1! 12C11 = 11 DVDs chosen at a time = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)(1) = 12

Check It Out: Example 3B Continued Course 3 10-9 Permutations and Combinations Check It Out: Example 3B Continued There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

Example How many different ways are there to purchase 2 CDs, 3 cassettes, and 1 videotape if there are 7 CD titles, 5 cassette titles, and 3 videotape titles? Answer: (7C2)(5C3)(3C1)= 630 ways