Suppose an $n$-sided die is rolled $t$ times. What is the probability that each side comes up at least $k$ times ($nk\leq t$)?
I've tried several combinatoric approaches, but every time I get stuck counting outcomes multiple times. For example
$$\binom{t}{nk}\frac{(nk)!}{k!^n}$$
yields the number of ways of placing the $nk$ constraining rolls, but the remaining outcomes cannot be placed freely (a factor of $n^{t-nk}$) without duplicating configurations from other placements of the constraining ones.
Is there a different approach I should be looking at, or a well-known solution?
Nothing like a complete answer, just one specific case.
When $k=1$, we see that we are counting the set of "onto" functions $\{1,2,\dots,t\}\to\{1,2,\dots,n\}$. This is related to the Stirling numbers of the second kind, namely it is:
$$n!S(t,n) =\sum_{i=0}^n (-1)^{n-i}{n \choose i} i^t$$
The formula on the right can be proved using inclusion-exclusion.
The probability is thus:
$$\frac{n!S(t,n)}{n^t} = \sum_{i=0}^{n}(-1)^{n-i}\binom{n}i\left(\frac in\right)^t$$
The inclusion-exclusion formula cannot be easily adjusted to deal with other values of $k$.