# 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 $$\displaystyle \binom{60}{10}$$

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.

• Why are we counting to $61$? There are only $60$ students.
– JMP
Commented Feb 15, 2023 at 6:46
• @JMP Hi, Actually that was the OP who considered $60$ students. I adopted a different approach, picking a $51-$element set out of $\{1,2, .., 61 \}$, and then taking the $11$th largest member of the chosen set into consideration, which is a trick to deal with the identity. Commented Feb 15, 2023 at 7:17

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

• I'm not the OP, but I was trying to come up with a solution and there's no way in any universe I'd ever get to this answer. Can you explain how you arrived at it please? Was this an identity you knew? Is there I can read about it? Or did you create it on your own? If so, where do I get such powers... Commented Feb 15, 2023 at 4:44
• @BigBear See math.stackexchange.com/questions/3506383/… Commented Feb 15, 2023 at 4:57

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,

$$[x^{10}]:\color{blue}{(1+x)^{40}}\color{fuchsia}{(1+x)^{20}}$$

$$=\sum_\limits{r=0}^{10} [x^{r}]:\color{blue}{(1+x)^{40}}[x^{10-r}]:\color{fuchsia}{(1+x)^{20}}$$

$$=\sum_\limits{r=0}^{10} \binom{40}{r}\binom{20}{10-r}$$

• Can you please tell me where is i am wrong. Commented Feb 15, 2023 at 14:04
• @PritiBisht; The expression you are evaluating isn't the same as this one.
– JMP
Commented Feb 15, 2023 at 14:05