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?

A bijection is a combinatorial way IMO. But there's a simple inductive argument. It's true for $n=1;$ suppose true for $n$ and take a set $S$ of size $n+1.$ Let $a$ be any element of $S$. Now $S- a$ is a set of size $n$, so has the same number of odd and even subsets. Therefore $S$ has the same number of odd subsets not containing $a$ as it has even subsets not containing $a.$ By adding $a$ to each subset, it also has the same number of even subsets containing $a$ as odd subsets containing $a.$ So in total it has the same number of odd and even subsets.
