Double Sum

This Problem is inspired by the "Point Path Probability" https://brilliant.org/mathematics-problem/point-path-probability

Let's show $\sum_{i=0}^{n}\sum_{j=0}^{n}\frac{{n \choose i} {n \choose j}}{{2n \choose i+j}} = 2n+1$ Rearranging, we have: $\sum_{i=0}^{n}\sum_{j=0}^{n}\frac{{n \choose i} {n \choose j}}{{2n \choose i+j}} = \sum_{k=0}^{2n}\frac{1}{{2n \choose k}}\sum_{i+j=k}{n \choose i} {n \choose j} = \sum_{k=0}^{2n}\frac{1}{{2n \choose k}}{2n \choose k} = 2n+1$ Where we applied Vandermonde's Identity above.

The second method is using double counting the pairs (s,p) where the point s is on the path p in the "Point Path Probability" problem: $\sum_{i=0}^{n}\sum_{j=0}^{n}{i+j \choose i} {2n-(i+j) \choose n-i} = (2n+1){2n\choose n}$ This is equivalent to the first identity by a simple algebraic manipulation.

Let $n=100$ , the answer to the problem is $201$ .