Simplification of $\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}$ There was a post on this web site an hour ago asking for the sum of
\begin{equation*}
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}
\end{equation*}
expressed as a single binomial coefficient. (Four choices were provided in the post.) The post seems to have been deleted.  I think it is worth keeping on this web site.

Solution
For any positive integer $n$ and any nonnegative integer $r \leq n$,
\begin{equation*}
\binom{n}{r} = \binom{n}{n - r} .
\end{equation*}
According to Vandermonde's Identity,
\begin{align*}
&\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\
&\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\
&\qquad = \sum_{r=0}^{49} \binom{50}{r} \binom{50}{49-r} \\
&\qquad = \binom{100}{49} .
\end{align*}
