Learning Intentions: h. 10.4 ounting Technique Use a tree diagram to represent possible paths or choices. Learn the definitions of & notations for permutations & combinations, & distinguish between them. Use the counting principle to count arrangements. ount permutations, combinations, or other arrangements to determine probabilities.
h. 10.4 ounting Techniques Practice Problems #2.) Evaluate each number of permutations or combinations without using your calculator. a. 6P 3 b. 6 3 c. 6 P 1 d. 6 1 #4.) Ricardo has six books written by five different authors. In how many ways can he arrange them on a shelf so that the two books by the same author are next to each other?
SOLUTIONS: h. 10.4 ounting Techniques Practice Problems #2.) Evaluate each number of permutations or combinations without using your calculator. a. 6P 3 b. 6 3 c. 6 P 1 d. 6 1 np r = 6P 3 = n r! 6! 6 3! 6P 3 = 6 5 4 3 2 1 3 2 1 6P 3 = 120 n r = 6 3 = n r! r! 6! 6 3! 3! 6 3 = 6 5 4 3 2 1 3 2 1 (3 2 1) 6 3 = 6 5 4 6 = 20 np r = 6P 1 = n r! 6! 6 1! 6P 1 = 6 5 4 3 2 1 5 4 3 2 1 6P 1 = 6 n r = 6 1 = n r! r! 6! 6 1! 1! 6 1 = 6 5 4 3 2 1 5 4 3 2 1 (1) 6 1 = 1
SOLUTIONS: h. 10.4 ounting Techniques Practice Problems #4.) Ricardo has six books (1-6) written by five different authors (A-E). In how many ways can he arrange them on a shelf so that the two books by the same author are next to each other? A 1 A 2 3 4 D 5 E 6 3 A 1 A 2 4 D 5 E 6... how many ways can a group be arranged in a set? Possible arrangements of groups within a set (X) for # of grouped items (g) given (n) total positions: If order of grouped items matters If order of grouped items does NOT matter X = [n g + 1] g! X = [n g + 1] X = [6 2 + 1] 2! X = [6 2 + 1] X = [5] (2 1) X = 5 (if book order of A 1 & A 2 X = 10 did not matter) However, the total number (T) of arrangements with grouping restriction = (# of ways remaining items can be arranged) (# of ways group can be arranged) T = (n g+1) P (n g) g! T = (6 2+1) P (6 2) 2! T = 5P 4 2! T = 5 4 3 2 2! Additional Reasoning: Think of the two books by the same author as stuck together so that there are 5 things to arrange. This makes 5! or 120 possibilities. For each of the 120 arrangements, the 2 books by the same author can be in 2! arrangements, giving (120)2, or 240.
ounting Techniques Suppose you take your little brother or sister out to a fastfood restaurant for dinner. The kids meal allows him to choose between four burgers, three drinks and two desserts. Give each item a name (& abbreviation) & list all possible meal combinations. How many possibilities are there?
Solutions: ounting Techniques Suppose you take your little brother or sister out to a fastfood restaurant for dinner. The kids meal allows him to choose between four burgers, three drinks and two desserts. How many different meal possibilities are there? Tree diagrams help to organize the different options and lists all the possible combinations. Meal #1 Meal #2 Meal #3 Meal #4 24 meals
Permutation: an arrangement of possible outcomes where order is important. Once a selection (choice) has been made the choice cannot be used again. np r = n r! np r = the number of permutations of n things chosen r at a time ex.) You have 5 pictures to arrange in a row along one wall, how many different orders can you place them? 5P 5 = 5 4 3 2 1 = 5 4 3 2 1 = 5! = 120 different orders order 1 st 2 nd 3 rd 4 th 5 th (if 1 out of the 5 pictures is hung, it options cannot be hung in the 2 nd, 3 rd, & so on, places) ex.) If there are 5 pictures to hang, but only 3 hooks how many different orders can you place the pictures in? Let n = 5 & r = 3 np r = 5P 3 = n r! 5! 5 3! 5P 3 = 5 4 3 2 1 2 1 5P 3 = 60
ombination: an arrangement in which the order is not important. Once a selection (choice) has been made the choice cannot be used again. n r = the number of combinations of n things chosen r at a time ex.) You have 6 coins: dollar, ½ dollar, quarter, dime, nickel & penny. If you shake the bank & three coins fall out, how many different sets of coins can you get? n r = 6 3 = 6 P 3 3P 3 = 120 6 6P 3 = 6 5 4 = 120 total sets of 3 (with repeats) 3P 3 = 3 2 1 = 6 (each combo. of 3 repeats 6x each) OR... = 20 i.e.) Dime, Nickel, Penny combo s = {DNP, DPN, NDP, NPD, PDN, PND} n r = 6 3 = n r! r! 6! 6 3! 3! 6 3 = 6 5 4 3 2 1 3 2 1 (3 2 1) 6 3 = 6 5 4 6 = 20 What is the probability you will get exactly 40 cents? P(40 cents) = # of combinations totally 40 = 1 total combinations 20 = 0.05 = 5% 40 = 0.25 + 0.10 + 0.05 (no other combo. = 40 )
Permutations, ombinations or neither? 1. At a restaurant, you select three different side dishes from eight possibilities. 2. The number of different committees of 10 students that can be chosen from the 50 members of the freshman class. 3. The number of different ice-cream cones if all three scoops are different flavors and a cone with vanilla, strawberry, then chocolate is different from a cone with vanilla, chocolate then strawberry. 4. The number of different three-scoop icecream cones if you can choose multiple scoops of the same flavor. 31 Flavors Ice ream Parlor
Solutions: Permutations, ombinations or neither? 1. At a restaurant, you select three different side dishes from eight possibilities. ombination; b/c the order does not matter & no dish can be chosen more than once. n r = 8 3 = 8 P 3 = 56 3 2 1 2. The number of different committees of 10 students that can be chosen from the 50 members of the freshman class. ombination; n r = 50 10 = 50P 10 3P 3 = 8 7 6 10 P 10 = 50 49 48 47 46 45 44 43 42 41 10! = 10,272,278,170 3. The number of different ice-cream cones if all three scoops are different flavors & a cone with vanilla, strawberry, then chocolate is different from a cone with vanilla, chocolate then strawberry. Permutation; n P r = 31 P 3 = 31 30 29 = 26,970 4. The number of different three-scoop ice-cream cones if you can choose multiple scoops of the same flavor. Neither, because repeating flavors is allowed. If 31 flavors total, then by the Multiplication ounting Rule: 31 31 31 = 29,791
versus Draw a tree diagram showing the possible results for a three game series between the Milwaukee rewers and the hicago ubs. Highlight the path indicating that the rewers won the first two games and the ubs won the final game. Does your diagram model permutations, combinations or neither? Explain. If each outcome is equally likely, what is the probability that the rewers won the first two games and the ubs the third? If you know that the rewers win more than one game, what is the probability that the sequence is?
Solutions: Draw a tree diagram showing the possible results for a three game series between the Milwaukee rewers and the hicago ubs. Highlight the path indicating that the rewers won the first two games and the ubs won the final game. () Does your diagram model permutations, combinations or neither? Neither. Repetition of team name is allowed. If each outcome is equally likely, what is the probability that the rewers won the first two games and the ubs the third? P() = 1 = 0.125 = 12.5% 8 If you know that the rewers win more than one game, what is the probability that the sequence is?) P() = = 0.25 = 25% # of games win > 1 = 1 4 Possible Outcomes of > 1: {,, & } Game 1 Game 2 Game 3 versus