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Prove that for any positive integer $n $: $(2n)!$ is divisible by $({n!}) ^2: $

If you know a solution to the problem, please help..

Sorry, this problem was edited by another user who changed some values. I have now fixed the problem.

Jmaxmanblue
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  • I don't get your question. $n!$ is not necessarily a perfect square, e.g. $4!=24$ in between $4^2$ and $5^2$. – GNUSupporter 8964民主女神 地下教會 Dec 09 '17 at 21:57
  • I dont understand the question. So you want to say for example for n=2 that 4!=24 is divisible by sqrt(2)? LOL, 3 same comments – croraf Dec 09 '17 at 21:57
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    From the very original version of the question, I believe that you want to say $(n!)^2$ divides $(2n)!$---but this is simply the binomial coefficient $\binom{2n}{n}$, which is an integer. – GNUSupporter 8964民主女神 地下教會 Dec 09 '17 at 21:59
  • $n!$ is never a square for $n>1$, so that doesn't make sense. –  Dec 09 '17 at 22:02
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    Your original version was "2 in (n!)2 is actually a square root", so don't blame others, please. –  Dec 09 '17 at 22:06
  • $\binom{2n}{n} $ is the simple answer to this. otherwise , you need to use the formula of checking the exponent of a prime $p$ in both $ 2n !$ and $n!$ using the De Polignac's Formula and show that the former is atleast two times more than the latter one. – Chirantan Chowdhury Dec 09 '17 at 22:07
  • To display square roots, use \sqrt. If the argument is more than one character enclose it in braces. So \sqrt {100} gives $\sqrt {100}$ – Ross Millikan Dec 09 '17 at 22:07

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From the comments $\frac {(2n)}{(n!)^2}={2n \choose n}$, which is guaranteed to be an integer.

Ross Millikan
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