We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+1 **

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