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.
 A: 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*}
A: There is another proof for this problem that I am mentioning here :
Since :
$$
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\
\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$
Consider the product :
$$(1+x)^{50} \times (1+x)^{50}$$
Look for the coefficient of $x^{49}$ in this product, it will be calculated as : 
$$ x^0 ~~\text{from first bracket} , x^{49} ~\text{from second bracket} = \binom{50}{0} \times  \binom{50}{49}$$
$$ x^1 ~~\text{from first bracket} , x^{48} ~\text{from second bracket} = \binom{50}{1} \times  \binom{50}{48}$$
$$ x^2 ~~\text{from first bracket} , x^{47} ~\text{from second bracket} = \binom{50}{2} \times  \binom{50}{47}
\\.
\\.
\\.
\\.$$
$$ x^{49} ~~\text{from first bracket} , x^{0} \text{from second bracket} = \binom{50}{49} \times  \binom{50}{0}$$
So, coefficient of $x^{49}$ in expansion of $(1+x)^{50}(1+x)^{50} =$
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$ 
Which is  coefficient of $x^{49}$ in expansion of $(1+x)^{100} =$
$$\binom{100}{49}$$
Thus :
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0}=\binom{100}{49}$$
