TITLE: Proof that binomial coefficients are integers - combinatorial interpretation
QUESTION [5 upvotes]: For any integers $k \le n$ here is an injective group homomorphism $$S_k \times S_{n-k} \rightarrow S_n$$ such that a tuple $(\sigma, \tau)$ permutes $\{1,...,n\}$ by letting $\sigma$ act on $\{1,...,k\}$ and $\tau$ act on $\{k+1,...,n\}$.
By Lagrange's theorem, $k!(n-k)! = |S_k \times S_{n-k}|$ divides $|S_n| = n!$, so $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is an integer.
In the comments of some question from several months ago (that I can't find now) it was asked if we can find a combinatorial interpretation of this result - that the cosets of $S_k \times S_{n-k}$ in $S_n$ should correspond naturally to ways of choosing $k$ elements from a set of $n$.
To me there is no obvious correspondence and I would like to know if anyone has an interpretation in that sense.

REPLY [4 votes]: Let $X$ be the set of $k$ element subsets of $\{1, \ldots, n\}$.  $S_n$ acts on $X$ (applying a permutation to $k$ distinct numbers yields another $k$ distinct numbers) and it should be clear that this action is transitive.  Thus the orbit stabilizer relation says that for any choice $x \in X$ there is a natural bijection $X \simeq S_n/\operatorname{Stab}(x)$ where $\operatorname{Stab}(x)$ are the permutations in $S_n$ that fix $x$.
If we let $x = \{1, \ldots, k\} \in X$ then $\operatorname{Stab}(x) = S_k \times S_{n - k}$ therefore the $X \simeq S_n/(S_k \times S_{n - k})$.