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Problem: Prove for all natural numbers the following identity:

$$\sum_{r=0}^{n}\frac{(2n)!}{(r!)^2((n-r)!)^2}=\dbinom{2n}{n}^2$$

I have just been successful in interpreting the LHS of the above as sum of the coefficients of those terms in the expansion of $(a+b+c+d)^{2n}$ which are of $(ab)^r(cd)^{n-r}$ form.

I also tried wrote the LHS in terms of binomial coefficients as follows

$$\sum_{r=0}^{n} \dbinom{2n}{r}\dbinom{2n-r}{r}\dbinom{2n-2r}{n-r}$$

But since $r+r+n-r$ is not a constant so the above sum cannot be interpreted as the coefficient of $x^{t}$ in some expression where t is some constant.

So, please help me with this problem. Even hints would be appreciated.

Also, I failed to find any combinatorial interpretation for this problem, though I am used to using double counting in combinatorial indentities.

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Rewrite $\frac{(2n)!}{(r!)^2((n-r)!)^2} = \binom{2n}{n}\binom{n}{r}\binom{n}{n-r}$. $$\therefore \sum_{r=0}^{n}\frac{(2n)!}{(r!)^2((n-r)!)^2}= \sum_{r=0}^{n}\binom{2n}{n}\binom{n}{r}\binom{n}{n-r} = \binom{2n}{n}\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r}$$ Now, $\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r}$ is just choosing $n$ objects out of total $2n$ objects. Thus, $\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r} = \binom{2n}{n}$ $$\therefore \frac{(2n)!}{(r!)^2((n-r)!)^2} = \binom{2n}{n}\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r} = \binom{2n}{n}^2$$

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