Question

How do you simplify `((2n)!)/(n!)`?

Answer 1

There are other ways of writing it, however none of them are simplifications.

Explanation:

While there isn't a simplification of `((2n)!)/(n!)`, there are other ways of expressing it. For example

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`((2n)!)/(n!) = prod_(k=0)^(n-1)(2n-k) = (2n)(2n-1)...(n+1)`

This follows directly from the definition of the factorial function and canceling common factors from the numerator and denominator.

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`((2n)!)/(n!) = 2^nprod_(k=0)^(n-1)(2k+1) = 2^n(1*3*5*...*(2n-1))`

A short proof of the identity:
`((2n)!)/(n!) = 1/(n!)(1*2*3*...*2n)`

`=1/(n!)*(2*4*6*...*2n) (1*3*5*...*(2n-1))`

`=1/(n!)(1*2)(2*2)(3*2)...(n*2)(1*3*5*...*(2n-1))`

`=1/(n!)(1*2*3*...*n)2^n(1*3*5*...*(2n-1))`

`=1/(n!)n!*2^n(1*3*5*...*(2n-1))`

`=2^n(1*3*5*...*(2n-1))`

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What form is best to use depends on the situation, however the given form of

`((2n)!)/(n!)`

is the most concise.