Prove that every $n$-element set has the same number of subsets with even and odd cardinality without creating a bijection I need to prove that the number of even and odd subsets of a finite set if always equal. This is not a problem once we assume that $n$ is odd. Then, every "even" subset of the set defines in a unique way an "odd" subset and vice versa. 
The problem starts once we assume that $n$ is even. The argument used in the previous case no longer works. I read some answers on Stack, but each of them, at this point, constructed a bijection which looked strange to me.  Is there a combinatorial or algebraical way to prove this theorem?

Give each size-$k$ subset a score of $(-1)^k$. The result you want is that the sum of all subsets' scores is $0$. But we can factorise that sum as $(1+(-1))^n$, since each element is either in or not in a subset, and multiplies the empty set's score of $1$ by $-1$ if it's included.
