TITLE: Probability that exactly k bins are empty, given m balls and n bins?
QUESTION [5 upvotes]: I've searched for an understandable answer to this exact question and have failed to find it.
How do you find the probability that exactly $k$ bins are empty, given $m$ balls and $n$ bins? (Each ball drop is independent).
The solution to this similar question does not explain how to find the probability that exactly $k$ bins are empty. It mentions the solution in passing in the comments, but does not thoroughly explain how to find the answer.

REPLY [4 votes]: (This is basically the same as true blue anil's answer) Let's count the ways in which $m$ distinguishable balls can be placed in $n$ bins.
Then the total number of possible ways is $n^m$
The "favorable" ways are those that have exactly $r=n-k\le m$ non-empty bins. There are $r! \, S(m,r) \, {n \choose r}$, where 

$r!$ counts the permutations of bins.
$S(m,r)$ is the Stirling number of the second kind, which count the number of ways of placing $m$ balls in $r$ unlabelled bins. 
${n \choose r}$ counts which bins are the empty (or non-empty) ones.

Then, because all positions are equiprobable, the desired probability is
$$P= \frac{r! \, S(m,r){n \choose r}}{n^m}=\frac{r! \, S(m,n-k){n \choose k}}{n^m}$$