Probability problem, a variant of the birthday problem

Question

There are $m$ bins and $n$ identical balls. We randomly (uniformly) throw the balls to the bins. Then we count the number of bins which only have 1 ball in it. What is the probability that this number is bigger than $t$, $(0 \leq t < m)$?

Answer 1

The number of combinations with exactly $k$ bins being occupied with a single ball is: $$\begin{align} N_k&=\binom nk \binom mk k!\sum_{i\ge0}(-1)^i\binom{n-k}i\binom{m-k}i i!(n-k-i)^{m-k-i}\\ &=\frac{ n! m!}{k!}\sum_{i\ge0}\frac{(-1)^i}{i!}\frac{(n-k-i)^{m-k-i}}{(n-k-i)!(m-k-i)!}, \end{align} $$ where $\binom nk$ and $\binom mk k!=\frac{m!}{(m-k)!}$ stay for the numbers of ways to choose $k$ bins to be occupied with a single ball and to place $k$ (distinguishable) balls into these bins, respectively. The following sum counts the number of ways to place the rest $m-k$ balls into $n-k$ bins with exclusion of the cases of a single ball placed in one of these bins (inclusion-exclusion principle). The factors $\binom{n-k}i$ and $\binom{m-k}i i!$ have here the same origin as discussed above.

Finally the probability in question is: $$ p_t=\frac1{n^m}\sum_{k>t} N_k. $$