cancelling out of $\frac{n!}{2n!}$? Perhaps a rather strange question. I am doing an exercise and the last part ends: 
$I = \frac{n!}{(2n)!}$
Now I know that: 
$$I = \frac{n(n-1)(n-2)(n-3)\cdots}{2n(2n-1)(2n-2)(2n-3)\cdots }$$
Is it possible to cancel out some terms? Or to make it a little bit more 'friendly'?
 A: I think your possible options could be summarised as:


*

*Leaving it as $\frac{n!}{(2n)!}$, which is a perfectly simple expression as it is.

*Canceling out the terms and expressing it as $\left[\prod_{i=1}^n(n+i)\right]^{-1}$, but this doesn't really achieve much.

*Using the Gamma function, to get $\frac{n!}{(2n)!}=\frac{\sqrt{π} }{2^{2 n}Γ(n + 0.5)}$. In a way this is simpler since the only hard part to compute is $Γ(n + 0.5)$. 

*Assuming that $n$ is sufficiently large and using $\frac{n!}{(2n)!}\sim \frac1{\sqrt{2}}\left(\frac{e}{4n}\right)^n $, by Stirling's approximation. In fact this approximation is pretty accurate, as shown for the true curve (blue), approximation (red) and absolute error (orange) below.

A: Note that $$I = \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2n(2n-1)(2n-2)(2n-3) \cdots 2 \cdot 1} $$$$= \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2^n \times n(2n-1)(n-1)(2n-3)(n-2) \cdots 1 \cdot 1}$$ and after cancelling out we get that$$ I = \frac{1}{2^n (2n-1) (2n-3) \cdots 3 \cdot 1} = \frac{1}{2^n} \prod_{i=1}^n \frac{1}{2i -1} $$This is one possible expression, but I think yours is more concise. 
A: Separating the even and odd terms, we have
$$ (2n)! = [ (2n)(2n-2)(2n-4)\dotsm \cdot 2 ] [(2n-1)(2n-3)\dotsm 1 ] \\
= 2^n [n(n-1)(n-2) \dotsm 1] [(2n-1)(2n-3)\dotsm 1] = 2^n n! (2n-1)!!, $$
where the double factorial is defined by $k!! = k(k-2)(k-4)\dotsm 3 \cdot 1 $ if $k$ is odd, or $\dotsm 4 \cdot 2$ if $k$ is even.
Alternatively, $(2n)!/n! = (2n)(2n-1)\dotsm (n+1) = (n+1)_{n}$, where $(a)_k$ is the rising factorial $ a(a+1)\dotsm (a+k-1) $.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
{n! \over \pars{2n}!} & =
{n\pars{n - 1}! \over \pars{2n}\pars{2n - 1}!} =
{1 \over 2}\,{\pars{n - 1}! \over \pars{2n - 1}!} =
{1 \over 2}\,{\Gamma\pars{n} \over \Gamma\pars{2n}}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & = \require{cancel}
{1 \over 2}\,{\cancel{\Gamma\pars{n}} \over \pars{2\pi}^{-1/2}
\,2^{2n - 1/2}\,\cancel{\Gamma\pars{n}}\Gamma\pars{n + 1/2}}\qquad
\pars{\begin{array}{l}
\Gamma\mbox{-}\,Duplication\ Formula\ \mbox{in the}
\\
\textsf{denominator}. 
\end{array}}
\\[5mm] & =
\bbx{{2^{-2n} \over \pars{n - 1/2}!}\,\root{\pi}} \\ &
\end{align}
A: Using the Double Factorial,
$$
(2n)!=(2n)!!(2n-1)!!=2^nn!(2n-1)!!
$$
Therefore,
$$
\frac{n!}{(2n)!}=\frac1{2^n(2n-1)!!}
$$
which does not really look any simpler.
A: Like @jam said you can use Stirling's approximation
This approximates to : $$\frac {2^{\frac{-4n-1}{2}}e^{n}}{n^n}$$
A: To complete the "panorama", consider also the expression in terms
of Rising ($x^{\,\overline {\,n\,}}$) and Falling ($x^{\,\underline {\, \,n\,}}$) Factorials
$$
{{n!} \over {\left( {2n} \right)!}} = {{1^{\,\overline {\,n\,} } } \over {1^{\,\overline {\,2n\,} } }} 
 = {{1^{\,\overline {\,n\,} } } \over {1^{\,\overline {\,n\,} } \;\left( {n + 1} \right)^{\,\overline {\,n\,} } }}
 = {1 \over {\;\left( {n + 1} \right)^{\,\overline {\,n\,} } }} = n^{\,\underline {\, - \,n\,} } 
$$
