Can anyone explain it? 

MrCreative  Sep 12, 2018

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

Rom  Sep 16, 2018

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