Evaluation of $\sum_{i=0}^{100}\sum_{j=0}^{100}\binom{100}{i}\cdot \binom{100}{j}\cdot\binom{200}{i+j}^{-1}$ Evaluation of $\displaystyle \sum_{i=0}^{100}\sum_{j=0}^{100}\frac{\binom{100}{i}\cdot \binom{100}{j}}{\binom{200}{i+j}} = $
$\bf{My\; Try::}$ Let $\displaystyle i+j=n\;,$ Then Sum convert into
$\displaystyle\sum_{n=0}^{200} \sum_{i=0}^{100}\frac{\binom{100}{i}\cdot \binom{100}{n-i}}{\binom{200}{n}} = \sum_{n=0}^{200}\binom{200}{n}^{-1}\cdot \sum_{i=0}^{100}\binom{100}{i}\cdot \binom{100}{n-i}$
Now $\displaystyle \sum_{i=0}^{100}\binom{100}{i}\cdot \binom{100}{n-i}=$ coeff. of $x^{n}$ in $\displaystyle (1+x)^{200} = \binom{200}{n}$
So Our Sum is $\displaystyle = \sum_{n=0}^{200}\binom{200}{n}^{-1}\cdot \binom{200}{n} = 201$
My Question is can we solve it using Combinotarail argument, If yes then plz explain me
Thanks
 A: We know from the $\color{blue}{\text{Hypergeometric}}$ (my namesake!) Distribution and the Vandermonde Identity that 
$$\large\sum_{r=0}^n\frac{{\binom Kr}{\binom{N-K}{n-r}}}{\binom Nn}=1$$
which happens to be independent of $n$. 
Hence, summing the above $n=0$ to $N$ gives
$$\large\sum_{n=0}^N \sum_{r=0}^n\frac{{\binom Kr}{\binom{N-K}{n-r}}}{\binom Nn}=N+1$$
Putting $N=2m, K=m$ gives
$$\large\sum_{n=0}^{2m} \sum_{r=0}^n\frac{{\binom mr}{\binom{m}{n-r}}}{\binom {2m}n}=2m+1$$
When $m=100$, the summation becomes
$$\large\sum_{n=0}^{200} \sum_{r=0}^n\frac{{\binom {100}r}{\binom{100}{n-r}}}{\binom {200}n}=201$$
Putting $j=r$ and $i=n-r=n-j$ (i.e. $i+j=n$) gives
$$\large\sum_{i=0}^{100} \sum_{j=0}^{100}\frac{{\binom {100}i}{\binom{100}{j}}}{\binom{200}{i+j}}=201\qquad\blacksquare$$

Alternatively: 
Putting $j=r$ and $i=n-r=n-j$ (i.e. $i+j=n$) gives
$$\large\sum_{i=0}^{m} \sum_{j=0}^m\frac{{\binom mi}{\binom{m}{j}}}{\binom{2m}{i+j}}=2m+1$$
When $m=100$, the summation becomes
$$\large\sum_{i=0}^{100} \sum_{j=0}^{100}\frac{{\binom {100}i}{\binom{100}{j}}}{\binom{200}{i+j}}=201\qquad\blacksquare$$
