Parity proof with binomial coefficients

There is only one condition required for a vector to be even: an even number of entries are odd.

For a solid example, consider vectors with four entries, $(x_1,x_2,x_3,x_4)$ . The total number of those vectors is $4^4=256$ . Then, any even vector can have 0, 2, or 4 odd entries. Here are the cases:

0 odd entries: $\displaystyle \binom{4}{0}=1$ arrangements

2 odd entries: $\displaystyle \binom{4}{2}=6$ arrangements

4 odd entries: $\displaystyle \binom{4}{4}=1$ arrangements

But take note that there are some degrees of freedom with each situation. For example, in the case of 0 odd entries, there are no choices for odd entries, but 2 choices for even. Therefore, there are $2^4=16$ choices of elements. Similarly, for 2 odd entries, there are $2^2\cdot 2^2=16$ choices, and for 4 odd entries, there are $2^4=16$ choices.

In summary, there are

$\displaystyle \binom{4}{0}\cdot 2^4+\binom{4}{2}\cdot (2^2\cdot 2^2)+\binom{4}{4}\cdot 2^4=128$

even vectors. This is exactly half of the total number of vectors.

Try to generalize this argument for the case where $n$ is arbitrary.