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When $n$ is a natural number, prove that $\frac{(2n)!}{n!^2}$ is a natural number.

If possible, I would like you to prove this by induction.

I tried to prove this by induction, but I can’t because $k+1$ is left in the denominator when I substitute $k+1$ to $n$.

Help me to solve this.

Gymnast
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Doesn't this suffice? $$ \frac{(2n)!}{n!^2} =\frac{(2n)!}{n!\,n!} =\binom{2n}{n} $$

lhf
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  • Still you have to prove that it is an integer: https://math.stackexchange.com/questions/382787/induction-proof-dbinom2nn-dfrac2nnn-is-an-integer?noredirect=1&lq=1 – Guy Fsone Dec 06 '17 at 12:14
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    @GuyFsone, not if you use that $\binom{m}{n}$ is the number of subsets of size $n$ out of $m$ objects. – lhf Dec 06 '17 at 12:17