+6252 Neat problem.
First off note that \(\dbinom{n}{k}=\dbinom{n}{n-k}\)
We are choosing (n-1) items from 2n possible items. One way to do this is to split the 2n items into 2 piles of n items each, choose k items from the first pile, and (n-1-k) items from the second pile. We do this for \(0 \leq k \leq n-1\)
This gets us
\(\dbinom{2n}{n-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{n-k-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{k+1}\)
you can recognize the right hand term as the summation shown
