sample2 soln

# sample2 soln - AMS 301 Sample Exam 2 Solution Ning SUN...

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Unformatted text preview: AMS 301 Sample Exam 2 Solution Ning SUN August 6, 2009 1. (20 pt) Consider the problem of counting the ways to distribute 29 identical objects into 6 boxes with at least 4 objects in each box. (a) (6 pt) Model this problem as an integer-solution-of-an-equation problem. x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 29 , x i ≥ 4 . (b) (7 pt) Model this problem as a certain coe cient of a generating function. ( x 4 + x 5 + x 6 + ··· ) 6 , we want to know the coe cient of x 29 (c) (7 pt) Solve this problem. Method 1: Give 4 objects to each box, and then distribute the remaining 29- 4 × 6 = 5 objects without restriction: C ( n + r- 1 ,r ) = C (6 + 5- 1 , 5) = C (10 , 5) = 10! 5!5! . Method 2: Using the generating function: ( x 4 + x 5 + x 6 + ··· ) 6 = x 4 (1 + x + x 2 + ··· ) 6 = x 24 (1 + x + x 2 + ··· ) 6 = x 24 ( 1 1- x ) 6 . So the coe cient of x 29 in ( x 4 + x 5 + x 6 + ··· ) 6 is the same as the coe cient of x 29- 24 = x 5 in ( 1 1- x ) 6 , which is ( 6+5- 1 5 ) = ( 10 5 ) = 10! 5!5! . 2. (20 pt) What is the probability that a 7-card poker hand chosen from the 52 cards in a deck has exactly 3 pairs (no 3-of-a-kind or 4-of-a-kind)? # of desired outcomes: Choose 3 kinds rst ( C (13 , 3) ways), for each of the three chosen kinds, choose a pair ( C (4 , 2) · C (4 , 2) · C (4 , 2) ways), then choose the last card from the remaining 40 cards (52-the three chosen kinds = 52- 12 = 40 ). So in total, # of ways= C (13 , 3) · [ C (4 , 2)] 3 · 40 = 13!...
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## This note was uploaded on 11/29/2011 for the course AMS 301 taught by Professor Arkin during the Spring '08 term at SUNY Stony Brook.

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sample2 soln - AMS 301 Sample Exam 2 Solution Ning SUN...

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