Evaluating $\binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{10}+\cdots+\binom{50}{10}\binom{10}{10}$ 
Finding value of
$$\binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{10}+\cdots+\binom{50}{10}\binom{10}{10}$$

Using $\displaystyle \binom{n}{r}=\binom{n}{n-r}$
$\displaystyle \binom{40}{0}\binom{20}{10}+\binom{41}{1}\binom{19}{9}+\cdots +\binom{50}{10}\binom{10}{0}$
I have tried it using combinational argument
Suppose we have $60$ students in a class in which $40$ boys and $20$ girls and we have to select $10$ students in which $r$ are boys and $n-r$ are girls
So total ways $\displaystyle \binom{60}{10}$
But my answer did not match with right answer
please have a look

$$
\sum_{k=0}^n \binom{4n+k}{k}\binom{2n-k}{n}
= \sum_{k=0}^n \binom{4n+k}{4n}\binom{2n-k}{n}
= \binom{6n+1}{5n+1} = \binom{6n+1}{n}
$$
Now take $n=10$ to obtain $\binom{61}{10}$.
