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

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

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

marked as duplicate by lab bhattacharjee, Brian M. Scott combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 23 '15 at 6:26

• 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... – Steven Stadnicki Jun 23 '15 at 6:16
• For $~n=\dfrac12~$ we have $~\displaystyle\sum_{n=0}^\infty\frac{\displaystyle{2n\choose n}^2}{16^n~(2n-1)^2}=\frac4\pi$ – Lucian Jun 23 '15 at 6:28