Prove $\binom{n}{0}$ + $\binom{n}{2}$ + $\binom{n}{4} + ...+\binom{n}{n-1}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...+\binom{n}{n}$ I want to prove that $\binom{n}{0}$ + $\binom{n}{2}$ + $\binom{n}{4} + ...+\binom{n}{n-1}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...+\binom{n}{n}$ for all odd numbers $n \ge 1$
I see that $\binom{n}{0}$ and $\binom{n}{n}$ are both equal to $1$ and therefor remove each other.
But what can I do after that? I'm very lost, please help me move in the right direction at least.

Here's an intuitive combinatorial proof.  The left-hand side is the number of even-sized subsets of an $n$-set, which we'll call $X$.  The right-hand side is the number of odd-sized subsets of $X$.  But $Y \mapsto X \setminus Y$ is a $1$-$1$ mapping of even-sized subsets onto odd-sized subsets, so the two numbers must be equal.
