This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 integersolutionofanequation 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 4card subset out of the 52card 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!...
View
Full
Document
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.
 Spring '08
 ARKIN

Click to edit the document details