If we define ${n \choose k}_m := \frac{n(n-m)(n-2m)...(n-(k-1)m)}{k!}$, then it can be shown that this is an integer value for all integer values of $n,k,m$. Does this function have a name? If ${n \choose k}$ counts the number of ways to choose $n$ objects from a collection of $k$, what does ${n \choose k}_m$ count?
EDIT: I've realized now that this is not an integer-valued function but in very particular circumstances.
Using Pochhammer symbols$${n \choose k}_m = \frac 1 {k!}\prod_{j=1}^k {(n-(j-1)\,m)}=\frac 1 {k!}(-m)^k \left(-\frac{n}{m}\right)_k$$ Using the gamma function $${n \choose k}_m =(-m)^k\,\frac{ \Gamma \left(k-\frac{n}{m}\right)}{\Gamma (k+1) \,\,\Gamma \left(-\frac{n}{m}\right)}$$
Up to today, there is no name for it (as far as I know) but it could be Feryll function tomorrow.
I am not aware of any existing name for this formula, but I can think of a scenario that gives the numerator of the formula.
Consider that we have $n$ objects (WLOG, $X = \{0, 1, 2, \ldots, n-1\}$) and we want to choose $k$ out of them. Suppose that whenever we choose one objective $x \in X$, $x$ and the $m - 1$ objects after $x$ (in a cyclic sense, i.e., if we reach the end then we continue to choose from the beginning) will all disappear and we cannot choose any of them in the following rounds, then the numerator of your formula (i.e., ${n \choose k}_m \times k! = n(n-m)(n-2m)...(n-(k-1)m)$) represents the possible ways to choose the $k$ objects considering the order.
However, we cannot simply divide it by $k!$ to have the unordered counterpart since possibly not all orderings of a group of $k$ objects are possible. For example, when $n = 5$, $k = 2$, and $m = 3$, we have the objects $$0, 1,2,3,4$$ and we can take $2$ first then take $0$, but we cannot take $0$ first then take $2$, since $0, 1, 2$ disappear together if we choose $0$ in the first round.
