exam2 solution

exam2 solution - AMS 301 Exam 2 Solution Ning SUN August...

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Unformatted text preview: AMS 301 Exam 2 Solution Ning SUN August 11, 2009 1. (20 pt) Consider the problem of counting the ways to select 25 objects from 8 types with at least 2 objects of each type. (a) (6 pt) Model this problem as an integer-solution-of-an-equation problem. x 1 + x 2 + + x 8 = 25 , x i 2 . (b) (7 pt) Model this problem as a certain coe cient of a generating function. ( x 2 + x 3 + x 4 + ) 8 , we want to know the coe cient of x 25 (c) (7 pt) Solve this problem. Method 1: Select 2 objects from each type, and then select the remaining 25- 2 8 = 9 objects without restriction: C ( n + r- 1 ,r ) = C (8 + 9- 1 , 9) = C (16 , 9) = 16! 9!7! . Method 2: Using the generating function: ( x 2 + x 3 + x 4 + ) 8 = x 2 (1 + x + x 2 + ) 8 = x 16 (1 + x + x 2 + ) 8 = x 16 ( 1 1- x ) 8 . So the coe cient of x 25 in ( x 2 + x 3 + x 4 + ) 8 is the same as the coe cient of x 25- 16 = x 9 in ( 1 1- x ) 8 , which is ( 8+9- 1 9 ) = ( 16 9 ) = 16! 9!7! . 2. (20 pt) What is the probability that a 4-card subset out of the 52-card deck has no pairs (each card has a di erent value)? # of desired outcomes: First choose 4 kinds from the 13 kinds: C (13 , 4) , and then choose one card from each kind C (4 , 1) 4 = 4 4 . So in total, the number is C (13 , 4) 4 4 = 13! 4 4 4!9!...
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This note was uploaded on 08/30/2010 for the course AMS 301 taught by Professor Arkin during the Spring '08 term at SUNY Stony Brook.

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exam2 solution - AMS 301 Exam 2 Solution Ning SUN August...

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