0
$\begingroup$

How do we show that this identity holds for any n? Any hints or solutions?

Show $\sum_{ r = 0}^{n} {(\binom n r)}^{2} = \frac{(2n)!}{(n!)^{2}}$

$\endgroup$

marked as duplicate by lab bhattacharjee, Brian M. Scott combinatorics Jun 23 '15 at 6:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Edited question. $\endgroup$ – islamfaisal Jun 23 '15 at 6:10
  • $\begingroup$ A broad hint: write one of your two $n\choose r$ terms as $n\choose n-r$ using the symmetry of the binomial coefficients. From there you should be able to apply familiar identities... $\endgroup$ – Steven Stadnicki Jun 23 '15 at 6:16
  • $\begingroup$ Already tried this but still stuck, can you give more hints? $\endgroup$ – islamfaisal Jun 23 '15 at 6:18
  • $\begingroup$ See Vandermonde's identity. $\endgroup$ – Lucian Jun 23 '15 at 6:23
  • $\begingroup$ For $~n=\dfrac12~$ we have $~\displaystyle\sum_{n=0}^\infty\frac{\displaystyle{2n\choose n}^2}{16^n~(2n-1)^2}=\frac4\pi$ $\endgroup$ – Lucian Jun 23 '15 at 6:28