hw1s - MS&E 120 Probabilistic Analysis Fall 2008-09 Prof...

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MS&E 120: Probabilistic Analysis Fall 2008-09, Prof. Shachter HOMEWORK #1 SOLUTIONS CHAPTER #1 PROBLEMS 3) 20 workers assigned to 20 jobs. Order the jobs in a particular list. Each ordering of workers assigned to this list is a different assignment. There are 20! ways of arranging the workers in a list. Therefore, # of possible assignments = 20!. 7) Ordering of boys and girls in a row. (a) There are 6 people in total (3 boys and 3 girls). # of ways of arranging them in a row = 6! = 720. (b) The boys and girls must each sit together in a group. Either the boys can be placed first or the girls can be placed first. For each of these 2 configurations, the boys can be arranged themselves in 3! ways (and similarly for the girls). # of ways = 2*3!*3! = 72. (c) The boys must sit together, so treat the boys as one individual person. Now these 4 people (the 3 girls and the group of boys) can be arranged in 4! ways. Once this order has been assigned (i.e. the position of the girls and the group of boys has been determined), the boys can be arranged in 3! ways among themselves. # of ways = 4!*3! = 144. (d) No two people of the same sex can sit together. The configuration can be of the type BGBGBG or GBGBGB, where B represents a boy and G a girl. The argument is similar to that in part (b). There are 2 basic configurations, and the boys and girls can be arranged separately in 3! ways each. # of ways = 2*3!*3! = 72. 8) Letter arrangements.

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MS&E 120: Probabilistic Analysis Fall 2008-09, Prof. Shachter This is a case of the number of permutations for n objects, given n 1 similar objects, etc. See Example 3d in the textbook. (a) FLUKE # of ways = 5! = 120. (b) PROPOSE # of ways = 7!/(2!*2!) = 1260. (c) MISSISSIPPI # of ways = 11!/(4!*4!*2!) = 34650. (d) ARRANGE # of ways = 7!/(2!*2!) = 1260. 12) 5 awards presented to 30 students. (a) Each of the 5 awards can go to any of the 30 students. This is equivalent to placing n=5 distinct balls into r=30 distinct urns. # of ways = 30 5 . (b) Each student can get a maximum of one award, and the awards are distinct. The first award can go to any of the 30 students, the second to any of the remaining 29, and so on… # of ways = (30)*(29)*(28)*(27)*(26). 13) The number of handshakes = number of pairs out of 20, since everyone shakes hands with each other. Note that the ordering of the pair is not important, i.e. Debarun shaking hands with Caleb is the same as Caleb shaking hands with Debarun! # of ways = 20 2 = 190. 14) Poker hand. There are 52 cards in a deck, and we can choose any 5 of them for a 5-card poker hand. # of ways = 52 5
MS&E 120: Probabilistic Analysis Fall 2008-09, Prof. Shachter 15) We can first choose the 5 women out of 10 and 5 men out of 12. Now we can place the women in a line, and then arrange the men in 5! ways to pair the men and women together. We have assumed that the order of the pairs is unimportant (i.e. if there is a

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This note was uploaded on 10/14/2008 for the course MS&E 120 taught by Professor Shatcher during the Fall '08 term at Stanford.

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hw1s - MS&E 120 Probabilistic Analysis Fall 2008-09 Prof...

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