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There are two sections on this homework exercise I'm stuck on. I'll appreciate some help!

We have 8 different boxes and 15 different glasses, divided between them. 3. What is the probability exactly two boxes will remain empty? 4. What is the probability the first 4 boxes would have an equal number of glasses?

For 3., I figured you have $C(8,6)$ ways to choose the non-empty boxes, and then you can multiply that by the number of ways to divide 15 different glasses in 6 boxes such that no box is empty. However, I don't know how many ways there are to do that (I only know the formula when the glasses are equivalent)

EDIT regarding 3.: Earlier today I had asked this question Ways to put numbered balls in boxes no box being empty, which has since gotten an answer. Using that I can solve 3. now. I'm still interested if there's an easier calculation, though!

For 4., I'm quite lost. I tried a number of methods, none of them seem to work. Only progress I really made is, you need to divide the question into cases: 1 glass in each of the first 4 boxes, 2 in each, and 3 in each.

I'd appreciate any help I can get! Thanks a lot.

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    I thought so as well, but I wrote a computer program that simulates this process and oddly enough, the results were quite different!2011-11-13
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    Edit: this comment was in response to a comment saying there was no difference whether the glasses are distinguishable or not. I think the author probably deleted the comment, or otherwise it somehow disappeared from sight.2011-11-13
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    Regarding $3$: You are computing the probability exactly two boxes have no balls. The colour of the balls makes no difference to this probability. Imagining them coloured, or not coloured, may help in the counting.2011-11-13
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    Hmm, but they are still different physical objects. So I think although you can think of them as combinatorially indistinguishable, the probability is nonetheless dependent on them being somehow different. (For example, two different colored dice have the same chance of getting (3,4) whether a color blind man sees them or not...)2011-11-13

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I assume problem (4) allows you to have empty boxes and that all boxes and glasses are distinguishable.

First choose how many glasses each of the first four boxes will get. You have 15 glasses total, so they can each have $i$, where $0 \leq i \leq 3$. Then, you choose four subsets of $i$ glasses, and put those subsets in the boxes in one of $4!$ ways. Then you have to distribute the remaining $15-4i$ glasses among the other four boxes, possibly allowing some of those four to be empty. For each of the remaining glasses you simply choose one of the four other boxes. So the number of ways to distribute the glasses is. $$ \sum_{i=0}^3 \binom{15}{i}\binom{15-i}{i}\binom{15-2i}{i}\binom{15-3i}{i}4!(4^{15-4i})$$ To get the probability, divide this by the total number of ways to distribute the glasses, which is simply $8^{15}$.

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    The first four boxes are supposed to have equal numbers of glasses. Everything is fine except the $4!$.2011-11-13
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    @Andre: I just realized you're right. The four binom's I have choose glasses for the first box, second box, etc. I think without the 4! it's OK.2011-11-13
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    Thanks for the answer! I have counted by hand that your solution (without the 4!) is correct for a small case (4 glasses, 3 boxes, 2 equal in size), but my program seems to disagree with you in the large case. The reasoning seems solid, so perhaps there's an error in the program, though.2011-11-13
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    @abcinmore What 'cases' are you referring to? What is your program trying to compute?2011-11-13