$\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof: How am I supposed to prove combinatorially:
$$\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$$
$${n\choose{0}}+{n\choose{2}}+{n\choose{4}}+\dots={n\choose{1}}+{n\choose{3}}+{n\choose{5}}+\cdots$$
Absolutely clueless.

The question as currently posed can be answered by looking at the symmetry of the  rows of Pascal's triangle corresponding to odd $n$ (which have an even number of elements).  By definition
$\large{n\choose{k}\large}=\frac{n!}{k!(n-k)!}$.
Therefore ${n\choose{0}}={n\choose{n}}$, ${n\choose{1}}={n\choose{n-1}}$, and in general ${n\choose{k}}={n\choose{n-k}}$.  Thus, the set of odd indexed elements and the set of even indexed elements in each row are identical.
