Combinatorial proof that the number of even cardinality subsets is equal to the number of odd cardinality subsets

Question

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?

Answer 1

Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $a\notin X$ and delete it if $a\in X$. One gets a pairing of the subsets, and in each pair one subset is even, the other odd.

Answer 2

use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.