# HW7 - HW7 Solutions Dilip Raghavan 1 Section 6.4 Problem 12...

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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 co-e ffi cient of x 5 5! should be 5 + 5 1 + 5 2 instead of just 1. Doing this with all the co-e 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 non–empty 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!...
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HW7 - HW7 Solutions Dilip Raghavan 1 Section 6.4 Problem 12...

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