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total:. pencil, highlighter, GP notebook, textbook, calculator. U11D5. Have out:. Bellwork:. 1. Compute the following without using a calculator (until the very last step). a). 10 P 3. b). 73 P 5.
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total: pencil, highlighter, GP notebook, textbook, calculator U11D5 Have out: Bellwork: 1. Compute the following without using a calculator (until the very last step). a) 10P3 b) 73P5 2. The Math Club is holding elections for next year’s leadership. There are 203 members, but only 4 positions are available (President, VP, Treasurer, and Secretary). How many ways can members be selected to fill the positions?
total: 1. Compute the following without using a calculator. a) 10P3 b) 73P5 +1 +1 +1 +1 +1 = 720 +1 = 1,802,440,080 2. The Math Club is holding elections for next year’s leadership. There are 203 members, but only 4 positions are available (President, VP, Treasurer, and Secretary). How many ways can members be selected to fill the positions? 203P4 = = 1,648,441,200 +1 +1 +1
Back to the movies: • Four friends go to the movies. In how many ways can they arrange themselves if… a) there are no restrictions? 4 3 2 1 ___ ___ ___ ___ = 4! = 24
Back to the Movies • Four friends go to the movies. In how many ways can they arrange themselves if… b) there are two couples? Remember: Even though there are 4 people, first focus on the “groups.” How many groups are there? 2 2 1 2! How many ways can 2 groups be arranged? ______ = How many ways can couple #1 be arranged within themselves? 2! How many ways can couple #2 be arranged within themselves? 2! Put it all together: 2! 2! 2! = 8
Back to the Movies • Four friends go to the movies. In how many ways can they arrange themselves if… c) there is 1 couple? How many groups are there? 3 couple, loner #1, loner #2 How many ways can 3 groups be arranged? 3 2 1 ______ ___ = 3! How many ways can the couple be arranged within themselves? 2! Put it all together: 2! 3! = 12
Back to the Movies • Four friends go to the movies. In how many ways can they arrange themselves if… d) two of the “friends” hate each other and cannot sit together? First: how many ways can the “frenemies” be seated? H1 H2 _________ ___ 3 ways H1 H2 _________ ___ H1 H2 _________ ___ Second: how many ways can the “frenemies” be arranged within themselves? 2! Third: how many ways can the others be arranged within themselves? 2! Put it all together: 2! 2! = 12 3
Back to the Movies • Six friends go to the movies. In how many ways can they arrange themselves if… a) there are 3 boys and 3 girls, and they must sit next to the opposite gender? How many groups are there? 2 How many ways can the groups be arranged? 2! _________ ___ ___ ___ _________ ___ ___ ___ B G B G B G G B G B G B How many ways can the boys be arranged? 3! How many ways can the girls be arranged? 3! Put it all together: 3! 3! 2! = 72
Back to the Movies • Six friends go to the movies. In how many ways can they arrange themselves if… b) there are 3 boys and 3 girls, and the boys are all together and the girls are all together? How many groups are there? 2 How many ways can the groups be arranged? 2! _________ ___ ___ ___ _________ ___ ___ ___ B B B G G G G G G B B B How many ways can the boys be arranged? 3! How many ways can the girls be arranged? 3! Put it all together: 3! 3! 2! = 72
3. Solve each problem. • The letters A, B, C, and D are used to form four–letter passwords for entering a computer file. How many passwords are possible if letters can be repeated? Order matters (because it’s a password!), repetition allowed 4 4 4 4 44 = 256 _____ _____ _____ _____ = b) How many five–digit numbers can be formed using the digits 4, 6, 7, 2, 8, 3, 5 if digits can NOT be repeated? Order matters, repetition NOT allowed, PERMUTATION!!! 7 6 5 4 3 2520 ____ ____ ____ ____ ____ = Write this as a permutation since we are selecting and arranging 5 things from a group of 7.
3. Solve each problem. c) Carlos has homework to do in math, chemistry, and English. How many ways can he choose the order in which to do his homework? Order matters, repetition NOT allowed, PERMUTATION!!! 3 2 1 _____ _____ _____ = 3! = 6 Write this as a permutation since we are selecting and arranging 3 things from a group of 3. Uh oh! What is 0! ???? Use your calculator to see the answer. We will look at “why” later.
3. Solve each problem. d) Lance’s math quiz has eight true–false questions. How many different choices for giving answers to the eight questions are possible? Order matters (because it’s a quiz!), repetition allowed 2 2 2 2 2 2 2 2 ___ ___ ___ ___ ___ ___ ___ ___ = 28 = 256 e) How many license plate numbers consisting of three letters followed by three numbers are possible when repetition is allowed? Order matters, repetition allowed 26 26 26 10 10 10 = 17,576,000 ___ ___ ___ ___ ___ ___ = 263● 103
3. Solve each problem. f) How many ways can nine different books be arranged on a shelf? Order matters, repetition NOT allowed, PERMUTATION!!! 9 8 7 6 5 4 3 2 1 ___ ___ ___ ___ ___ ___ ___ ___ ___ = 9! = 362,880 Write this as a permutation since we are selecting and arranging 9 books from a group of 9 books.
3. Solve each problem. g) Sixteen teams are competing in a soccer match. Gold, silver, and bronze medals will be awarded to the top three finishers. In how many ways can the medals be awarded? Order matters, repetition NOT allowed, PERMUTATION!!! 16 15 14 ___ ___ ___ =
3. Solve each problem. h) How many license plate numbers consists of three letters followed by three numbers are possible when repetition is NOT allowed? Order matters, repetition NOT allowed, PERMUTATION!!! 26 25 24 10 9 8 11,232,000 ___ ___ ___ ___ ___ ___ = This problem can be written as two permutations. The first permutations is selecting and arranging 3 letters from a group of 26. The second permutations is selecting and arranging 3 numbers from a group of 10.
4. Use a calculator to evaluate each expression. = 1 = 1 a) 0! b) 1! 5. Evaluate each expression without a calculator. Show your work! 5 4 3 2! 102 101 100! b) a) 2! 3! 100! = 5 4 3 = 60 = 1717
6. Evaluate each expression without a calculator. Show your work! a) 5P2 = 5 • 4 = 20 7. How many different arrangements can be made with the letters in the last names of these Presidents? a) EISENHOWER b) KENNEDY 3 E’s 2 E’s and 2 N’s = 604,800 = 1260
