The function which returns the number of ways to choose k items from n items without repetition and without order was added inissue35431. This functions is always goes in pair with other function, which returns the number of ways to choose k items from n items without repetition and with order. These functions are always learned together in curses of combinatorics. Often C(n,k) is determined via P(n,k) (and both are determined via factorial).    P(n, k) = n! / (n-k)!    C(n, k) = P(n, k) / k!The proposed PR adds meth.perm(). It shares most of the code with math.comb().

+0 from me.  It is inevitable that this will be requested now that we have comb().  The need for this is much less though.

> +0 from me.  It is inevitable that this will be requested now that we have comb().Exactly my thoughts, too. I don't think I'll have any use for `math.perm`; the main reason to add it will be to satisfy all those who got taught combinations and permutations at the same time, see them as a unit, and then get surprised when they find `math.comb` without `math.perm`.The good part is that now that we have math.comb, I think everything about the API of math.perm is already clear (including what to call it).

New changeset5ae299ac78abb628803ab7dee0997364547f5cc8 by Serhiy Storchaka in branch 'master':bpo-37128: Add math.perm(). (GH-13731)https://github.com/python/cpython/commit/5ae299ac78abb628803ab7dee0997364547f5cc8