Evaluating $\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$ The question is to evaluate $$\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$$I tried rewriting the expression as $$\sum_{r=0}^{10}\frac{\binom{20-r}{10}\binom{30+r}{r}}{\binom{50}{30}\binom{20}{10}}$$ I am facing trouble on how to further simplify the numerator.Any ideas?Thanks.

$$\begin{align}
\sum_{r=0}^{10}\frac {\displaystyle\binom {10}r}{\displaystyle\binom{50}{30+r}}
&=\frac {\color{blue}{\displaystyle\sum_{r=0}^{10}\displaystyle\binom {20-r}{10}\binom {30+r}{30}}}{\color{}{\displaystyle\binom {50}{20}\binom {20}{10}}}\\\\
&=\frac {\color{blue}{\displaystyle\binom {51}{41}}}{\color{green}{\displaystyle\binom {50}{20}\binom {20}{10}}}
&&\scriptsize\color{blue}{\text{using }\sum_{r=0}^{a-b}\binom {a-r}b\binom {c+r}d=\binom {a+c+1}{b+d+1}}\\\\
&=\frac {\displaystyle\binom {51}{10}}{\color{green}{\displaystyle\binom {50}{10}\binom {40}{10}}}
&&\scriptsize \color{green}{\text{using }\color{green}{\binom ab\binom bc=\binom ac\binom {a-c}{b-c}}}
\\\\
&= \frac {51\cdot 10!\;30!}{41!}\\\\
&=\color{red}{\frac 3{2\; 044\; 357\; 744}}\\\\
\end{align}$$
See also wolframalpha confirmation here and here.

NB: The first line makes use of the following:
$$\begin{align}
\binom {20}{10}\binom {10}r&=\binom {20}r\binom {20-r}{10}
&&\scriptsize\text{using }\binom ab\binom bc=\binom ac\binom {a-c}{b-c}\\
\color{orange}{\binom {50}{20}}\binom {20}{10}\binom {10}r&=\color{orange}{\binom {50}{20}}\binom {20}r\binom {20-r}{10}
&&\scriptsize\text{multiplying by }\binom {50}{20}\\
\color{purple}{\binom{50}{10}\binom{40}{10}}\binom{10}r
&=\binom {50}{20}\binom {20}r\binom{20-r}{10}
&&\scriptsize\text{using }\binom ab\binom bc=\binom ac\binom {a-c}{b-c}\\
&=\color{green}{\binom{50}{30+r}\binom{30+r}{30}}\binom{20-r}{10}
&&\scriptsize\text{using }\binom ab\binom {a-b}c=\binom a{b+c}\binom {b+c}{b}\\
\frac{\displaystyle\binom{10}r}{\displaystyle\binom{50}{30+r}}&=
\frac {\displaystyle\binom{20-r}{10}\binom{30+r}{30}}{\displaystyle\binom {50}{10}\binom {40}{10}}
\end{align}$$
