If $n$ balls are randomly placed into $n$ cells, find the probability that exactly one cell remains empty.
So the denominator is $n^n$. I can pick the empty cell in $n$ ways. There are $n-1$ cells that can have $k$ balls. There are $\binom{n}{k}$ ways of picking the $k$ balls to go into this cell. Also there are $(n-k) \cdots (n-2)$ ways of placing the remaining $n-k$ balls into the $n-2$ cells. So the product of this is $n(n-1) \binom{n}{k} [(n-k) \cdots (n-2)]$
So the answer is $\frac{n(n-1) \binom{n}{k} [(n-k) \cdots (n-2)]}{n^n}$
Is this right?