For a positive integers n, prove that

$$\displaystyle\sum\limits_{v=0}^n \frac{(2n)!}{(v!)^2 ((n-v)!)^2} = \binom{2n}{n}^2.$$

If somebody could please help me with this question, I would greatly appreciated it.

  • $\begingroup$ what do you mean by $\ast$? Multiplication? $\endgroup$ – TZakrevskiy May 14 '13 at 10:01
  • $\begingroup$ I've corrected your formula. $\endgroup$ – azimut May 14 '13 at 10:09

A combinatorial proof. The right-hand side corresponds to choosing two (possibly overlapping) sets of size $n$ from $2n$ elements.

Now we observe that:

$$\dfrac{2n!}{v!^2(n-v)!^2} = \binom{2n}{n} \binom n v \binom n{n-v}$$

If we divide the set of $2n$ elements in two fixed sets of $n$ elements, then picking $n$ from the original set amounts to picking $v$ from the first set, and $n-v$ from the second, for $v =0 \ldots n$.

Thus we find that

$$\sum_{v=0}^n \binom nv \binom n{n-v} = \binom{2n}n$$

which finally leads to the desired

$$\sum_{v=0}^n \frac{2n!}{v!^2(n-v)!^2} = \sum_{v=0}^n\binom{2n}{n} \binom n v \binom n{n-v} = \binom{2n}n \sum_{v=0}^n \binom nv\binom n{n-v} = \binom{2n}n^2$$


$$ \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\frac{(2n)!}{n!n!}\cdot \left(\frac{n!}{v!(n-v)!}\right)^2=\binom{2n}n \cdot \left(\frac{n!}{v!(n-v)!}\right)^2$$

$$\sum\limits_{v=0}^n= \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\binom{2n}n \cdot\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)^2$$

Now, equate the coefficients of $x^n$ in the following identity $$(1+x)^n(x+1)^n=(1+x)^{2n}$$ to find $$\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)\cdot \left(\frac{n!}{v!(n-v)!}\right)=\binom{2n}n$$


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