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
 Spring '08
 ARKIN
 1L, Vowel, 2m, 6 pt, 7 PT

Click to edit the document details