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Given $K$ balls and $M$ buckets and the limits $L1$ and $L2$ where $0<L1<L2<M$. We distribute all balls in the buckets randomly so a bucket can end up with $0$ to $K$ balls.

How do I calculate the number of combinations that have a number of empty buckets $E$ where $L1<E<L2$?

As I understand it without the restrictions the number of combinations are $$ (K+M-1)!/(K!(M-1)!) $$ but I do not know where to go from here.

If it makes it easier we can add the restriction $K<=M$ or even $K=M$

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    $\begingroup$ There appears to be a clash between "$K$ balls" and "...with $0$ to $M$ balls". $\endgroup$ – Rebecca J. Stones Dec 28 '17 at 4:29
  • $\begingroup$ Thanks, my bad, I fixed it $\endgroup$ – JJ Doe Dec 28 '17 at 13:37
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It looks like the bins are distinguishable. In this case, if we know which $E$ buckets are empty, we can use stars and bars (the positive integer version). So we sum over the possible $E$-subsets: $$ \sum_{\substack{\mathcal{E} \subseteq \{1,\ldots,M\} \\ L1 < |\mathcal{E}| < L2}} \binom{K-1}{M-|\mathcal{E}|-1} = \sum_{E=L1+1}^{L2-1} \binom{M}{E} \binom{K-1}{M-E-1}. $$

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  • $\begingroup$ It works, thanks! $\endgroup$ – JJ Doe Dec 28 '17 at 16:37

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