Combinatorial proof of a binomial identity: $\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$ I am to find a combinatorial argument for the following identity:
$$\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$$
For the right hand side, I was think that would just be number of ways to choose at least $s-1$ elements out of a $[2r-s]$ set. However, for the left hand side, I don't really know what it is representing. 
Any help would be greatly appreciated!

Suppose we seek to verify that
$$\sum_{k=1}^r {2r\choose 2k-1} {k-1\choose s-1}
= 2^{2r-2s+1} {2r-s\choose s-1}$$
where presumably  $s\ge 1$.  The lower  limit is set  to $k=1$  as the
first binomial coefficient is zero when $k=0.$
Introduce
$${2r\choose 2k-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{2r-2k+2}} \frac{1}{(1-z)^{2k}}
\; dz.$$
This provides range control and vanishes when $k\gt r$
so we may extend the range to infinity, obtaining
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} 
\sum_{k\ge 1} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz.$$
This yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} 
\sum_{k\ge s} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}}
\sum_{k\ge 0} {k+s-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}}
\frac{1}{(1-z^2/(1-z)^2)^s}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}} 
\frac{1}{((1-z)^2-z^2)^s}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}} 
\frac{1}{(1-2z)^s}
\; dz.$$
This is
$$[z^{2r-2s+1}] \frac{1}{(1-2z)^s}
= 2^{2r-2s+1} {2r-2s+1+s-1\choose s-1}
\\ = 2^{2r-2s+1} {2r-s\choose s-1}$$
and we have the claim.
