I'm struggling with finding a non-recursive formula for calculating the following probability:
Probability of throwing exactly $V$ distinct sides (numbers) on $N$ sided dice by $K$ rolls
Example for a normal dice where number of sides $N=6$, rolls $K=6$ and required distinct sides $V=2$ can be found here: Six throws, only two distinct numbers: coincidence?
The formula in the example is not general and only solution I can find is recursive one which I cannot use.
This would appear to be using Stirling numbers of the second kind:
$$\frac{1}{N^K} \times {N\choose V} \times V! \times {K\brace V}.$$
These probabilities sum to one (${K\brace 0} = 0$ when $K\ge 1$):
$$\frac{1}{N^K} \sum_{V=0}^N ·{N\choose V} V! K! [z^K] \frac{(\exp(z)-1)^V}{V!} \\ = \frac{1}{N^K} K! [z^K] \sum_{V=0}^N ·{N\choose V} V! \frac{(\exp(z)-1)^V}{V!} \\ = \frac{1}{N^K} K! [z^K] \sum_{V=0}^N ·{N\choose V} (\exp(z)-1)^V = \frac{1}{N^K} K! [z^K] \exp(Nz) = \frac{1}{N^K} N^K = 1.$$