Given a multiset $ S = \{\!\!\{ m \cdot 0, 1, ..., n \}\!\!\}$ where $m, n \in \mathbb{N}$, find the number of $m$-permutations of $S$.
For example, when $m=2$ and $n=2$, $S = \{\!\!\{ 0, 0, 1, 2 \}\!\!\}$ and the $2$-permutations are: $$ (0, 0), (0, 1), (0, 2), (1, 0), (1, 2), (2, 0), (2, 1)$$
Similar questions have been asked before ([1] and [2]), but no correct or concise close form solution was given in either case.
I prefer to use lower-case for the natural numbers here: $S=\{\!\!\{m\cdot0,1,2,\ldots,n\}\!\!\}$. As usual let $[n]=\{1,\ldots,n\}$.
Suppose that $0\le k\le m$; we’ll count the $m$-permutations of $S$ with $k$ zeroes. There are $\binom{m}k$ ways to choose the positions of the zeroes and $(m-k)!\binom{n}{m-k}$ ways to fill the remaining $m-k$ positions with an $(m-k)$-permutation of $[n]$, so there are
$$(m-k)!\binom{m}k\binom{n}{m-k}=(m-k)!\binom{m}{m-k}\binom{n}{n-(m-k)}$$
$m$-permutations of $S$ with $k$ zeroes. Thus, there are altogether
$$\sum_{k=0}^m(m-k)!\binom{m}{m-k}\binom{n}{n-(m-k)}=\sum_{k=0}^mk!\binom{m}k\binom{n}{n-k}=\sum_{k=0}^mk!\binom{m}k\binom{n}k$$
$m$-permutations of $S$. I don’t at the moment see any nice simplification of this sum.
