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: HW7 Solutions Dilip Raghavan October 25, 2008 1 Section 6.4 Problem 12. Let us first see what the generating function would be if we treated the first two boxes as one box. Since b 1 < b 2 4, the minimum value for b 1 + b 2 is 1 and the maximum value is 7. So if the first two boxes were combined together into one box, the generating function would be x 1! + x 2 2! + x 3 3! + x 4 4! + x 5 5! + x 6 6! + x 7 7! ! x 0! + x 1 1! + x 2 2! + ! 3 Now, if b 1 + b 2 is equal to, for example, 5, we need to figure out how many ways there are to distribute 5 distinct balls into boxes 1 and 2 with b 1 < b 2 4. This is the same as deciding which balls will go into box 1. We could have 0, 1 or 2 balls in box 1. So there are 5 + 5 1 + 5 2 ways to distribute 5 distinct balls into boxes 1 and 2 with b 1 < b 2 4. So in the first factor above, the coe ffi cient of x 5 5! should be 5 + 5 1 + 5 2 instead of just 1. Doing this with all the coe ffi cients, we get the exponential generating function to be 1 ! x 1! + 2 ! x 2 2! + " 3 ! + 3 1 !# x 3 3! + " 4 ! + 4 1 !# x 4 4! + " 5 ! + 5 1 ! + 5 2 !# x 5 5! + " 6 ! + 6 1 ! + 6 2 !# x 6 6! + + " 7 ! + 7 1 ! + 7 2 ! + 7 3 !# x 7 7! ! x 0! + x 1 1! + x 2 2! + ! 3 Problem 14. Let us fix n and m . Let a r be the number of ways to distribute r distinct balls into n distinct boxes with exactly m nonempty boxes. First consider the exponential generating function for the number of ways to distribute r distinct balls into m distinct boxes with no box empty. This is given by x 1!...
View
Full
Document
 Spring '11
 Hernandez/Rag

Click to edit the document details