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

I think more clarification is needed although @RobPratt has already given a solution. So, I am going to post a more detailed solution.

Let's consider the sum in a different way.
Assume we want to choose $51$ numbers from the set $\{1,2, ..,61\}$. Now, consider the $11$th largest number among those chosen. It must be at most $51$ and at least $41$.
If the $11$th largest number of the chosen set is $51$, we have $\binom{10}{10} \binom {50}{40}$ options.
If the $11$th largest number of the chosen set is $50$, we have $\binom{11}{10} \binom {49}{40}$ options;
and similarly, the same goes for the rest, in particular, if the $11$th largest number of the chosen set is $41$, we have $\binom{20}{10} \binom {40}{40}$ options.
Therefore:
$$\binom{61}{51}=\binom{10}{10}\binom{50}{40}+\binom{11}{10} \binom {49}{40}+ ... +\binom{20}{10} \binom {40}{40} \\= \binom{10}{10}\binom{50}{10}+\binom{11}{10} \binom {49}{9}+ ... +\binom{20}{10} \binom {40}{0}. \\$$
We are done.
