1. Warm Up Evaluate. 1. 5  4  3  2  1 2. 7  6  5  4  3  2  1 3.4. 5.6. 120 5040 4210 10 70

2. Sect. 11.1, 11.2

3. Solve problems involving the Fundamental Counting Principle. Solve problems involving permutations and combinations. Objectives Fundamental Counting Principle permutation factorial combination Vocabulary

4. You have previously used tree diagrams to find the number of possible combinations of a group of objects. In this lesson, you will learn to use the Fundamental Counting Principle.

5. Use a tree diagram Snowboarding A sporting goods store offers 3 types of snowboards (all mountain, freestyle, and carving) and 2 types of boots (soft and hybrid). How many choices does the store offer for snowboarding equipment? Draw a tree diagram and count the number of branches. The tree has 6 branches. So, there are 6 possible choices.

6. Using the Fundamental Counting Principle To make a yogurt parfait, you choose one flavor of yogurt, one fruit topping, and one nut topping. How many parfait choices are there? Yogurt Parfait (choose 1 of each) Flavor Plain Vanilla Fruit Peaches Strawberries Bananas Raspberries Blueberries Nuts Almonds Peanuts Walnuts number of flavors times number of fruits number of nuts times equals number of choices 2  5  3 = 30 There are 30 parfait choices.

7. Using the Fundamental Counting Principle A password for a site consists of 4 digits followed by 2 letters. The letters A and Z are not used, and each digit or letter many be used more than once. How many unique passwords are possible? digit digit digit digit letter letter 10  10  10  10  24  24 = 5,760,000 There are 5,760,000 possible passwords.

8. You try… A “make-your-own-adventure” story lets you choose 6 starting points, gives 4 plot choices, and then has 5 possible endings. How many adventures are there? number of starting points  number of plot choices number of possible endings  = number of adventures 6  4  5 = 120 There are 120 adventures.

9. You try… A password is 4 letters followed by 1 digit. Uppercase letters (A) and lowercase letters (a) may be used and are considered different. How many passwords are possible? Since both upper and lower case letters can be used, there are 52 possible letter choices. letter letter letter letter number 52  52  52  52  10 = 73,116,160 There are 73,116,160 possible passwords.

10. Use the fundamental counting principle Number of ways = 12 55 11 = 7260 SOLUTION You can use the fundamental counting principle to find the total number of ways to frame the picture. Multiply the number of frame styles (12), the number of frame colors (55), and the number of mat boards (11). Photography You are framing a picture. The frames are available in 12 different styles. Each style is available in 55 different colors. You also want blue mat board, which is available in 11 different shades of blue. How many different ways can you frame the picture?

11. Use the counting principle with repetition License Plates The standard configuration for a Texas license plate is 1 letter followed by 2 digits followed by 3 letters. How many different license plates are possible if letters and digits can be repeated? a. How many different license plates are possible if letters and digits cannot be repeated? b. There are 26 choices for each letter and 10 choices for each digit. You can use the fundamental counting principle to find the number of different plates. = 45,697,600 = 26 10 10 26 26 26 = 32,292,000 = 26 10 9 25 24 23

12. A permutation is a selection of a group of objects in which order is important. There is one way to arrange one item A. A second item B can be placed first or second. A third item C can be first, second, or third for each order above. 1 permutation 2 · 1 permutations 3 · 2 · 1 permutations You can see that the number of permutations of 3 items is 3 · 2 · 1. You can extend this to permutations of n items, which is n · (n – 1) · (n – 2) · (n – 3) ·... · 1 This expression is called n factorial, and is written as n!.

14. Sometimes you may not want to order an entire set of items. Suppose that you want to select and order 3 people from a group of 7. One way to find possible permutations is to use the Fundamental Counting Principle. First Person Second Person Third Person There are 7 people. You are choosing 3 of them in order. 7 choices 6 choices5 choices  = 210 permutations

15. Find the number of permutations Olympics Ten teams are competing in the final round of the Olympic four- person bobsledding competition. In how many different ways a. can the bobsledding teams finish the competition? (Assume there are no ties.) In how many different ways can 3 of the bobsledding teams finish first, second, and third to win the gold, silver, and bronze medals? b. There are 10! different ways that the teams can finish the competition. = 10 9 8 7 6 5 4 3 2 1 = 3,628,800 10 9 8 = 720

16. arrangements of 4 4! 4 · 3 · 2 · 1 Another way to find the possible permutations is to use factorials. You can divide the total number of arrangements by the number of arrangements that are not used. In the previous slide, there are 7 total people and 4 whose arrangements do not matter. arrangements of 7 = 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 210 This can be generalized as a formula, which is useful for large numbers of items.

18. Finding Permutations How many ways can a student government select a president, vice president, secretary, and treasurer from a group of 6 people? This is the equivalent of selecting and arranging 4 items from 6. = 6 5 4 3 = 360 Divide out common factors. There are 360 ways to select the 4 people. Substitute 6 for n and 4 for r in

19. Finding Permutations How many ways can a stylist arrange 5 of 8 vases from left to right in a store display? Divide out common factors. = 8 7 6 5 4 = 6720 There are 6720 ways that the vases can be arranged.

20. You try… Awards are given out at a costume party. How many ways can “most creative,” “silliest,” and “best” costume be awarded to 8 contestants if no one gets more than one award? = 8 7 6 = 336 There are 336 ways to arrange the awards.

21. Find the number of permutations. 5P35P3 ( 5 – 3 )! 5P35P3 = 5!5! = 60 = 5 4 3 2 1 2 1 4P14P1 ( 4 – 1 )! 4P14P1 = 4!4! = 4 = 4 3 2 1 2 1

22. Find permutations of n objects taken r at a time Music You are burning a demo CD for your band. Your band has 12 songs stored on your computer. However, you want to put only 4 songs on the demo CD. In how many orders can you burn 4 of the 12 songs onto the CD? SOLUTION Find the number of permutations of 12 objects taken 4 at a time. ( 12 – 4 )! 12 P 4 = 12! = 11,800 = 12! 8! 479,001,600 40320 = ANSWER You can burn 4 of the 12 songs in 11,880 different orders.

24. Find permutations with repetition Find the number of distinguishable permutations of the letters in a. MIAMI and b. TALLAHASSEE. SOLUTION MIAMI has 5 letters of which M and I are each repeated 2 times. So, the number of distinguishable permutations is a. = 30. 2! 5!5! = 120 2 TALLAHASSEE has 11 letters of which A is repeated 3 times, and L, S, and E are each repeated 2 times. So, the number of distinguishable permutations is b. = 831,600. 3! 2! 2! 2! 11! = 39, 916,800 6 2 2 2

25. Find the number of distinguishable permutations of the letters in the word. MALL MALL has 4 letters of which L repeated 2 times. So, the number of distinguishable permutations is = 12. = 4 3 2! 4!4! KAYAK KAYAK has 5 letters of which K and A are repeated 2 times. So, the number of distinguishable permutations is = 30. 2! 2! 5! 120 2 =

26. A combination is a grouping of items in which order does not matter. There are generally fewer ways to select items when order does not matter. For example, there are 6 ways to order 3 items, but they are all the same combination: 6 permutations  {ABC, ACB, BAC, BCA, CAB, CBA} 1 combination  {ABC} When deciding whether to use permutations or combinations, first decide whether order is important. Use a permutation if order matters and a combination if order does not matter.

27. To find the number of combinations, the formula for permutations can be modified. Because order does not matter, divide the number of permutations by the number of ways to arrange the selected items.

29. You can find permutations and combinations by using nPr and nCr, respectively, on scientific and graphing calculators. Helpful Hint

30. Application There are 12 different-colored cubes in a bag. How many ways can Randall draw a set of 4 cubes from the bag? Step 1 Determine whether the problem represents a permutation of combination. The order does not matter. The cubes may be drawn in any order. It is a combination. = 495 There are 495 ways to draw 4 cubes from 12. Step 2 Use the formula for combinations. 5

31. You try… The swim team has 8 swimmers. Two swimmers will be selected to swim in the first heat. How many ways can the swimmers be selected? = 28 The swimmers can be selected in 28 ways. 4

32. Find the number of combinations. 8C38C3 1. (8 – 3)! 3! = 8! 8C38C3 =56 SOLUTION 6 = 336 = 8 7 6 5! 5! 3! = 5010 24 =210 SOLUTION 2. 10 C 6 = 10 9 8 7 6! 4! 6! (10 – 6)! 6! = 10! 10 C 6

33. Find combinations Cards A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit. a. If the order in which the cards are dealt is not important, how many different 5- card hands are possible? b. In how many 5 -card hands are all 5 cards of the same color?

34. Find combinations SOLUTION The number of ways to choose 5 cards from a deck of 52 cards is: = 2,598,960. = 52 51 50 49 48 47! 47! 5! = 52! 52 C 5

35. Find combinations = 131,560. 21! 5! = 26! 26 C 52C12C1 1! 2!2! = 26 25 24 23 22 21! 21! 5! 1! 2 b. For all 5 cards to be the same color, you need to choose 1 of the 2 colors and then 5 of the 26 cards in that color. So, the number of possible hands is:

36. When finding the number of ways both an event A and an event B can occur, you need to multiply. When finding the number of ways that event A or event B can occur, you add.

37. Decide to multiply or add combinations Theater William Shakespeare wrote 38 plays that can be divided into three genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies. a. How many different sets of exactly 2 comedies and 1 tragedy can you read? b. How many different sets of at most 3 plays can you read?

38. Decide to multiply or add combinations SOLUTION a. You can choose 2 of the 18 comedies and 1 of the 10 tragedies. So, the number of possible sets of plays is: = 153 10 9! 1! = 10! 10 C 118 C 2 16! 2! 18! = 18 17 16! 16! 2 1 9! 1 10 9! = 1530

39. Decide to multiply or add combinations b. You can read 0, 1, 2, or 3 plays. Because there are 38 plays that can be chosen, the number of possible sets of plays is: 38 C 0 + 38 C 1 + 38 C 2 + 38 C 3 = 1 + 38 + 703 + 8436 = 9178

40. Counting problems that involve phrases like “at least” or “at most” are sometimes easier to solve by subtracting possibilities you do not want from the total number of possibilities.

41. Solve a multi-step problem Basketball During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend? SOLUTION Of the 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is: 12 C 3 + 12 C 4 + 12 C 5 +………..+ 12 C 12

42. Solve a multi-step problem Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 2 12 total combinations. If you attend at least 3 games, you do not attend only a total of 0, 1, or 2 games. So, the number of ways you can attend at least 3 games is: 2 12 – ( 12 C 0 + 12 C 1 + 12 C 2 ) = 4096 – (1 + 12 + 66) = 4017

43. Homework p.798: 9-15, 17-19, 21, 23, 31