HW6Answers(2.4)

HW6Answers(2.4) - Math331 Spring 2008 Instructor David Anderson Section 2.4 Homework Answers Homework pgs 63 66#’s 1 2 7 13 19 30 1 This is a

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 2.4 Homework Answers Homework: pgs. 63 - 66 #’s 1 , 2 , 7 , 13 , 19 , 30. 1. This is a straightforward application. We have ( 20 6 ) = 38 , 760. 2. There are 100 senators. To get a majority, there needs to be x > 50 senators. For each such x there are ( 100 x ) ways to achieve a majority. Therefore, there are N = 100 summationdisplay x =51 parenleftbigg 100 x parenrightbigg = 100 summationdisplay x =51 100! x !(100- x )! . The solution above is acceptable. There is a clearer answer, however. Note that the equation above also gives the total number of ways to get a minority of senators, or subsets from 0 to 49 (since every majority is associated with a unique minority). Thus, 2 N = total # of subsets- 50/50 tie = 2 100- parenleftbigg 100 50 parenrightbigg = 2 100- 100! (50!) 2 . So, N = 2 99- 100! 2 × (50!) 2 . Either of the solutions above are acceptable....
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This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at Wisconsin.

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HW6Answers(2.4) - Math331 Spring 2008 Instructor David Anderson Section 2.4 Homework Answers Homework pgs 63 66#’s 1 2 7 13 19 30 1 This is a

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