The probability of distributing $k$ balls over $n$ boxes with exactly $q$ boxes empty. I'm trying to solve this question and wound up with the following reasoning, but I'm not quite sure if it is correct.
The total number of possibilities of distribution $k$ balls over $n$ boxes can be given by $\binom{n+k-1}{n-1}$. Now we need the possibilities to distribute $k$ balls over $n-q$ boxes,$\binom{k+n-q-1}{n-q-1}$ multiplied by the possible picks of $q$ boxes out of $n$,$\binom{n}{q}$. Thus resulting in $\binom{k+n-q-1}{n-q-1}\binom{n}{q}$. 
The total probability of exactly $q$ boxes remaining empty is then given by:
$$P = \frac{\binom{k+n-q-1}{n-q-1}\binom{n}{q}}{\binom{n+k-1}{n-1}}.$$
Is this the correct way of calculating this. I'm mostly unsure about the numerator.
 A: You have not specified the random protocol according to which the balls are distributed over the boxes. I shall assume the following: One ball after the other is thrown into one of the $n$ boxes, uniformly (i.e., all boxes have equal probability) and independently. The result of the experiment is a word of length $k$ over the alphabet $[n]$, and all such words have the same probability.
You want to know the probability that the resulting word omits exactly $q$ letters from $[n]$. Therefore we have to count the number of such words. We can select the set $F$ of forbidden letters  in ${n\choose q}$ ways. When this set has been selected an admissible word is then a surjective map $f:\>[k]\to F':=[n]\setminus F$. Such an $f$  induces a partition of the set $[k]$ into $|F'|=n-q$ unlabeled nonempty blocks, which then can be assigned to the elements of $F'$ in $(n-q)!$ ways. The number of such partitions is the Stirling number of the second kind $S(k,n-q)$. In all, the number $N_{\rm adm}$ of admissible words then comes to
$$N_{\rm adm}={n\choose q}\>S(k,n-q)\>(n-q)!\ ,$$
and the requested probability $P$ is given by
$$P={N_{\rm adm}\over n^k}\ .$$
