<s>[INST] 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
 [/INST] 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,
$[x^{10}]:\color{blue}{(1+x)^{40}}\color{fuchsia}{(1+x)^{20}}$
$=\sum_\limits{r=0}^{10} [x^{r}]:\color{blue}{(1+x)^{40}}[x^{10-r}]:\color{fuchsia}{(1+x)^{20}}$
$=\sum_\limits{r=0}^{10} \binom{40}{r}\binom{20}{10-r}$
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