Evaluate:$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$ Evaluate:$$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$$
My Attempt:
I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.

Using the identity $$
\binom{51}{k}=\binom{50}{k}+\binom{50}{k-1},
$$ we get 
$$
\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}=\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}+\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}.$$ Observe that the former is equal to
$$
\sum_{k=0}^{50} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}= \sum_{ 0\leq r<k\leq 50}\binom{50}{k}\binom{50}{r} = \sum_{ 0\leq k<r\leq 50}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}.
$$ Also note that the latter is equal to
$$\begin{eqnarray}
\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}=\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}.
\end{eqnarray}$$ Gathering them together yields
$$
\sum_{k=0}^{50} \sum_{r=k+1}^{50}\binom{50}{k}\binom{50}{r}+\sum_{k=0}^{50}\sum_{r=0}^{k}\binom{50}{k}\binom{50}{r}=\sum_{k=0}^{50} \sum_{r=0}^{50}\binom{50}{k}\binom{50}{r} = 2^{50}\cdot2^{50} = 2^{100}.
$$ Hence the answer is $\frac{1}{2}$.
