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
 A: $$
\sum_{k=0}^n \binom{4n+k}{k}\binom{2n-k}{n}
= \sum_{k=0}^n \binom{4n+k}{4n}\binom{2n-k}{n}
= \binom{6n+1}{5n+1} = \binom{6n+1}{n}
$$
Now take $n=10$ to obtain $\binom{61}{10}$.
A: 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.
A: Your answer is wrong, you have evaluated the wrong sum, you have done this
There are $60$ students, $40$ boys and $20$ girls. This can be written as
$(1+x)^{60}=\color{blue}{(1+x)^{40}}\color{fuchsia}{(1+x)^{20}}$
As we wish to choose $10$ students, we want the coefficient of $x^{10}$, which is written $[x^{10}]:(1+x)^{60}$, and equals $\binom{60}{10}$.
On the RHS,
$[x^{10}]:\color{blue}{(1+x)^{40}}\color{fuchsia}{(1+x)^{20}}$
$=\sum_\limits{r=0}^{10} [x^{r}]:\color{blue}{(1+x)^{40}}[x^{10-r}]:\color{fuchsia}{(1+x)^{20}}$
$=\sum_\limits{r=0}^{10} \binom{40}{r}\binom{20}{10-r}$
