Given a set of cardinality $n\geq 1$, the number of subsets of even cardinality is equal to the number of subsets of odd cardinality
I am looking for a combinatorial proof of this statement- I know an algebraic proof is easy, for instance by expanding $(1-1)^n$. If $n$ is odd it is easy to see there is a one-to-one correspondence between even-cardinality subsets an odd-cardinality subsets using the complement. However, I can't think of a combinatorial proof if $n$ is even. Any ideas?