The title is pretty self-explanatory as to what my question is, I think.
If we have distributed $nk$ distinct elements among $n$ boxes such that each box has $k$ objects in it, and we then randomly redistribute the objects so that each box still has $k$ objects, what is the probability that the contents of every box have changed?
My first thought was to compute the number of ways of rearranging the elements in which no box has the same contents as before via inclusion exclusion. My work is as follows:
For each $i \in [n]$, there are $n \choose i$$\frac{(nk-ik)!}{(k!)^{n-i}}$ ways of rearranging so that there are at least $i$ boxes which remained the same.
The total number of ways of rearranging is $\frac{(nk)!}{(k!)^n}$, (which is the multinomial $nk \choose k, k, ..., k$).
So the probability that we want is:
$\sum_{i=0}^n {n \choose i}\frac{(nk-ik)!(k!)^i(-1)^i}{(nk)!}$
That is, unless I have made a mistake along the way. If my answer so far is correct, then here is the issue which I am having- I don't see how to simplify it further. Is it possible to find a closed form of the probability that we want here, or is my answer essentially as good as it gets?