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!
 A: HINT: We have $2r-s$ white balls numbered $1$ through $2r-s$. We pick $s-1$ of them and paint those balls red, and we stick gold stars on any subset of the remaining white balls; since there are $2r-s-(s-1)=2r-2s+1$ white balls remaining, there are $$\binom{2r-s}{s-1}2^{2r-2s+1}$$ possible outcomes of this sequence of operations.
Alternatively, we can start with $2r$ white balls numbered $1$ through $2r$. We pick an odd number of these balls, $2k-1$ for some $k\in[r]$, and line them up in numerical order. The $k$-th ball in line is the one in the middle; call it $B$. We choose $s-1$ of the $k-1$ chosen balls with numbers smaller than that of $B$ and paint them red. This is possible only if $k\ge s$, in which case the number on $B$ is at most $2r-s+1$, and the red balls all have numbers in the set $[2r-s]$. We now throw away the balls with numbers not in $[2r-s]$ and stick gold stars on any white balls left in the set of chosen balls. At this point we have $s-1$ red balls and possibly some white balls with gold stars. There are 
$$\sum_k\binom{2r}{2k-1}\binom{k-1}{s-1}$$
possible outcomes. 


*

*Verify that these possible outcomes are exactly the same as for the first sequence of operations.

A: 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.
A: $$
\begin{align}
\sum_k\binom{2r}{2k-1}\binom{k-1}{s-1}
&=\sum_k\binom{2r}{2r-2k+1}\binom{k-1}{k-s}\tag1\\
&=\sum_k(-1)^{k-s}\binom{2r}{2r-2k+1}\binom{-s}{k-s}\tag2\\[3pt]
&=\left[x^{2r-2s+1}\right](1+x)^{2r}\left(1-x^2\right)^{-s}\tag3\\[12pt]
&=\left[x^{2r-2s+1}\right](1+x)^{2r-s}(1-x)^{-s}\tag4\\[9pt]
&=\sum_k(-1)^k\binom{2r-s}{2r-2s-k+1}\binom{-s}{k}\tag5\\
&=\sum_k\binom{2r-s}{2r-2s-k+1}\binom{k+s-1}{k}\tag6\\
&=\sum_k\binom{2r-s}{k+s-1}\binom{k+s-1}{s-1}\tag7\\
&=\binom{2r-s}{s-1}\sum_k\binom{2r-2s+1}{k}\tag8\\
&=2^{2r-2s+1}\binom{2r-s}{s-1}\tag9
\end{align}
$$
Explanation:
$(1)$: symmetry of Pascal's triangle
$(2)$: apply negative binomial coefficient
$(3)$: write the sum as the coefficient in a product
$(4)$: cancel factors
$(5)$: write the coefficient in a product as a sum
$(6)$: apply negative binomial coefficient
$(7)$: symmetry of Pascal's triangle
$(8)$: $\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$
$(9)$: evaluate the sum
