I know that to distribute n identical presents to k children so that each child gets at least one is $\binom{n-1}{k-1}$ and that to distribute n presents to $k$ children is $\binom{n+k-1}{k-1}$, and I understand why. In order for each child to get at least $2$ presents is $\binom{n-k-1}{k-1}$. However I have no clue how to make it so a child doesn't receive more than a certain number. Any ideas? Thanks
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Im thinking its [number of ways to distribute presents to children with no restriction] - [number of ways to distribute presents to children such that at least 1 child gets 4+ presents] – Anon Mar 17 '15 at 22:40
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I'd go with your #no restrictions - #at least one child gets at least 4 presents. Are you having difficulty determining how to count the "bad" cases that you're subtracting off? – TravisJ Mar 18 '15 at 00:36
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You need to use an inclusion-exclusion argument. For a concrete example see this question and its answers; for the general result see this one. – Brian M. Scott Mar 18 '15 at 01:12
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possible duplicate of Number of ways to put $n$ unlabeled balls in $k$ bins with a max of $m$ balls in each bin – David K May 16 '15 at 04:42