summing this binomial series I found a really interesting question which is as follows:
Prove that the value of
$$\sum^{7}_{k=0}[({7\choose k}/{14\choose k})*\sum^{14}_{r=k}{r\choose k}{14\choose r}] = 6^7$$
my approach:
I tried to simplify the innermost sigma as well as  trying to simplify by using
${n\choose k}=n!/k!(n-k)!$ however I am can't get hold of this one.
My guess is that the summation simplifies into a standard series but I can't say for sure.
Kindly help me out.

Using $${14 \choose r}{r \choose k} = {14 \choose k}{14-k \choose r-k}$$ given reduces to
$$
\begin{align*}
  & \sum_{k=0}^7 {7 \choose k} \bigg\{\sum^{14}_{r=k} {14-k \choose r-k} \bigg\}  \\
  & = \sum_{k=0}^7 {7 \choose k} \{2^{14-k}\} \\
  & = 2^{7} \times \sum_{k=0}^7 {7 \choose k} 2^{7-k} \\
  & = 2^{7}\times(2+1)^{7} \\
  & = 6^7 
\end{align*}
$$
Edit : As pointed by @ElliotYu, outer bound should be from $0$ to $7$.
