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There are $\binom{r+n-1}{n-1}$ ways for distributing $r$ identical balls into $n$ distinct boxes. The $m$ part is throwing me off. I would say there are $\binom{r+n-m-1}{n-m-1}$ ways, but I'm not sure if I'm right.

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It's not quite correct. Think about it as a two-stage process:

  • choose the empty boxes: $\binom nm$ ways
  • put the $r$ balls into $n-m$ boxes: $\binom{r+n-m-1}{n-m-1}$ ways, as you wrote

So your answer is off, but only by a factor of $\binom nm$.

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enter image description here

See the above table with all the required combinations

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