This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: HW6 Solutions Dilip Raghavan October 19, 2008 1 Section 6.2 Problem 23. Let a n denote the number of ways to choose n boxes of candies with at least one but not more than 5 boxes of each type. The generating function for the sequence of a n is ( x + x 2 + x 3 + x 4 + x 5 ) 7 Since we want to choose 300 candies, we want to select 15 boxes of candy. So we want the coe ffi cient of x 15 in the above polynomial. Factoring out x , we have ( x + x 2 + x 3 + x 4 + x 5 ) 7 = x 7 h 1 + x + x 2 + x 3 + x 4 i 7 So we want the coe ffi cient of x 8 in h 1 + x + x 2 + x 3 + x 4 i 7 . We use the identity for 1 x 5 1 x h 1 + x + x 2 + x 3 + x 4 i 7 = " 1 x 5 1 x # 7 = 7 ! 7 1 ! x 5 + 7 2 ! x 10 7 2 ! x 15 + ··· ! + 7 1 ! + 1 + 7 1 1 ! x + 2 + 7 1 2 ! x 2 + ··· ! Since we want the coe ffi cient of x 8 , the answer is 7 ! 8 + 7 1 8 ! 7 1 ! 3 + 7 1 3 ! . Problem 26. Let us first count the number of ways to divide the objects into ordered plies. We can then divide that by 2 to get the answer. This works because the two piles cannot be the same. This is because there are an odd number of onjects of each kind. So it is not possible for both piles to have the same number of, say, apples. If there were an even number of objects of each kind, then we would have to consider the case when both piles were the same, and so we cannot just divide the answer by 2.case when both piles were the same, and so we cannot just divide the answer by 2....
View
Full
Document
This note was uploaded on 09/15/2011 for the course AMAT 356 taught by Professor Hernandez/rag during the Spring '11 term at SUNY Albany.
 Spring '11
 Hernandez/Rag

Click to edit the document details