Combinatorial interpretation of the identity: ${n \choose k} = {n \choose n-k}$ What is the combinatorial interpretation of the identity: ${n \choose k} = {n \choose n-k}$?
Proving this algebraically is trivial, but what exactly is the "symmetry" here. Could someone give me some sort of example to help my understanding? 
EDIT: Can someone present a combinatorial proof?

Let $S$ be a set of $n$ elements. Choosing $m$ members of $S$ leaves a remainder of $n-m$ members in $S$.  
Thus, each $m$-subset of $S$ corresponds to exactly one $(n-m)$-subset of $S$, and conversely, to any such $(n-m)$-subset, there corresponds exactly one $m$-subset of $S$.
Therefore, $\binom{n}{m}=\binom{n}{n-m}$.
