Prove that (2n)! is bigger than (n!)^2

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We prove a nice identity for positive integers #shorts

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MathElite
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(2n)!/(n!)^2 is just the number of ways of choosing n elements from a group of 2n. So of course it’s bigger than 1 for n>0. Then the equality follows.

federicovolpe
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And this is true for all the positive numbers including fractions except between 0.76 and 0.86. And also true for all the negative integers but not all the negative fractions.

mathevengers
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It's true for all Real greater than -0.5
:)

pardeepgarg