# HW5Answers(2.3) - (8(2 × 3 × 3 6 8 = 0 000333 15 There...

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Math331, Spring 2008 Instructor: David Anderson Section 2.3 Homework Answers Homework: pgs. 50 - 53 #’s 1 , 6 , 10 , 13 , 15. 1. There are 4! ways to sort the 4 items. Therefore, the probability of guessing correctly is 1 / 4! = 1 / 24 = 0 . 0417. 6. We need 6 P 2 = 6! / 4! = 30. He needs 30 one way dictionaries. 10. This is a direct application of Theorem 2.4. There are 11! / (3! × 2! × 3! × 3!) = 92 , 400 distinguishable outcomes. 13. There are N = 6 8 diFerent possible outcomes if one tosses a die eight times. Using Theorem 2.4, there are 8! / (2! × 3! × 3!) ways to get exactly two 3’s, three 1’s, and three 6’s. Therefore, the probability of such an occurrence is
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Unformatted text preview: (8! / (2! × 3! × 3!)) 6 8 = 0 . 000333 . 15. There are n + m drinks total. Assume that the men choose their drinks sequentially. There are n + m ways for the ±rst man to choose, n + m-1 for the second, . .., n + m-( n-1) = m + 1 for the n th. Therefore, there are ( n + m ) ··· ( m + 1) = ( n + m )! /m ! ways for the men to choose drinks. There is precisely one way that the men choose their drinks correctly. Thus, the probability of this happening is 1 / (( n + m )! /m !) = m ! / ( n + m )!. 1...
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