This might be a standard, simple question, and I am willing to research it but don't know where to start. I believe I know what the answer might be but can't prove it, or at least need a simple pointer.
Given $S_n$, the symmetric group on $[n]={1,..,n}$, consider two (non-empty) subsets $A,B$ of size $k<n$, both of which contain a given integer $j$. The question I'm researching is this: How many permutations (i.e. elements of $S_n$) take, under the action of $S_n$ induced on the subsets of $[n]$ of size $k$, $B$ onto $A$? My guess is that this does not depend on $A$ or $B$, only $k$ and, if so, then the answer is some multiple of $(k−1)!$ (in the case $A=B$) but, again, am not sure.
One way to attack this would be to simply observe that any permutation that ends up taking $B$ onto $A$ must be the composition of a single permutation doing so and a permutation of the $k−1$ elements of $A$, which would give $(k−1)!$ (the single, original permutation being the composition of itself with the identity). But I don't think I'm counting quite right as there are multiple elements of $S_n$ which, when restricted to subsets of size $k$, give the same mapping.
Thanks for any references.
Your permutation carries $B$ onto $A$, and also carries the complement of $B$ onto the complement of $A$. So you should get $k!(n-k)!$, by multiplying the number of possibilities for each of the two parts of the permutation, if $A,B$ both have size $k$. (note I put $k!$ instead of $(k-1)!$, because the number of bijections between two sets of size $k$ is $k!$)