Distribute $r$ indistinguishable balls randomly into $n$ cells.
The probability that exactly $m$ cells will contain exactly $k$ balls each is?
The answer: $$\frac { m!r!} { m!n^{r} } \begin{equation} \sum_{j}{\frac {(n−j)^{r−jk}} {(j−m)!(n−j)!(r−jk)!(k!)^{j}}} \end{equation} $$ When then sum is for $$ j≥m $$ s.t. $$ j≤n,kj≤r $$
I tried to get to the above phrase without success, it is really a difficult exercise.