TITLE: With $n$ balls and $n$ bins, what is the probability that exactly $k$ bins have exactly $1$ ball?
QUESTION [10 upvotes]: I've got a balls and bins problem.  Suppose I throw $n$ balls uniformly at random into $n$ bins.  What is the probability that exactly $k$ bins end up with exactly $1$ ball?
I know this seems a classical problem and may look "simple" or "naive," but I've worked days on it and still can't get the answer.
However, I think I do have a good approximation for it. Namely, let $X$ denote the number of such bins.  Then
$$ Pr(X=k) \approx \binom{n}{k}\left(\frac{1}{e}\right)^{k}\left(1-\frac{1}{e}\right)^{n-k} $$
where $1/e$ is an approximation for $(1-1/n)^{n-1}$.
This approximation works great when $n$ is big and poorly when $n$ is small (like $n<5$).
Anyway, I'm looking for an exact expression.  Anyone have an idea?
PS: I've written a simple simulation in C++; you can check your answer with it first: Simulation Code Here.

REPLY [6 votes]: Let $S_i$ be the set of outcomes with $1$ ball in bin $i$. Let $N_j$ be the number of outcomes in the intersections of $j$ of the $S_i$; e.g. $N_3=\sum_{i<j<k}|S_i\cap S_j\cap S_k|$. There are $\binom{n}{j}$ choices of the $S_i$ to intersect, for each choice of $S_i$, there are $\binom{n}{j}j!$ choices and orders of balls to put into those $j$ bins, and $(n-j)^{n-j}$ ways to arrange the other $n-j$ balls in the other $n-j$ bins. Thus,
$$
N_j=\binom{n}{j}\binom{n}{j}j!(n-j)^{n-j}
$$
Since there are $n^n$ possible outcomes, to compute the probability of getting exactly $k$ bins with $1$ ball, use the Generalized Inclusion-Exclusion Principle:
$$
\begin{align}
\frac{1}{n^n}\sum_{j=k}^n(-1)^{j-k}\binom{j}{k}N_j
&=\frac{1}{n^n}\sum_{j=k}^n(-1)^{j-k}\binom{j}{k}\binom{n}{j}\binom{n}{j}j!(n-j)^{n-j}\tag{1}\\
&=\frac{1}{n^n}\sum_{j=k}^n(-1)^{j-k}\binom{n}{k}\binom{n-k}{n-j}\frac{n!}{(n-j)!}(n-j)^{n-j}\\
&=\binom{n}{k}\frac{n!}{n^n}\sum_{j=k}^n(-1)^{j-k}\binom{n-k}{n-j}\frac{(n-j)^{n-j}}{(n-j)!}\\
&=\binom{n}{k}\frac{n!}{n^n}\sum_{j=0}^{n-k}(-1)^{n-k-j}\binom{n-k}{j}\frac{j^j}{j!}
\end{align}
$$
Appendix:
To verify that the probabilities for $k=0,1,\dots,n$ sum to $1$, $(1)$ can be summed fairly easily in $k$:
$$
\begin{align}
&\sum_{k=0}^n\frac{1}{n^n}\sum_{j=k}^n(-1)^{j-k}\binom{j}{k}\binom{n}{j}\binom{n}{j}j!(n-j)^{n-j}\\
&=\frac{1}{n^n}\sum_{j=0}^n\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}\binom{n}{j}\binom{n}{j}j!(n-j)^{n-j}\\
&=\frac{1}{n^n}\sum_{j=0}^n(-1)^{j}0^j\binom{n}{j}\binom{n}{j}j!(n-j)^{n-j}\\
&=\frac{1}{n^n}(-1)^00^0\binom{n}{0}\binom{n}{0}0!(n-0)^{n-0}\\
&=1
\end{align}
$$
Mathematica:
Here is the plot for $80$ bins, which matches Sasha's plot: