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

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?

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.
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.