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simplify (2n)!/(2n-3)!
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top of the fraction is 1 * 2 * 3 * ... * (2n-3) * (2n-2) * (2n-1) * (2n). bottom of the fraction is 1 * 2 * 3 * ... * (2n-3). all factors cancel, and you're left with (2n-2) * (2n-1) * (2n) on the top of the fraction.
\[{2n! \over (2n-3)!}={(2n-3)!(2n-2)(2n-1)(2n) \over (2n-3)!}=4n(n-1)(2n-1)\]
Thanks guys, I knew the answer, but I didn't know how I would get to the 4n part.
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