# Show as $n$ get large that $e^{\frac{-n^2}{2K}}$ is an approximation of $\frac{k!}{k^n(k-n)!}$

I'm asked to show this on a problem I'm working on.

$e^{\frac{-n^2}{2K}}$ is an approximation of $\frac{k!}{k^n(k-n)!}$ when $n$ is large.

However in this class we've never gone over $e$, so the only forumula I know is $e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$. I don't see how to use that. Is there another forumula or method I'm not see or aware of? Any help is greatly appreciated. Thanks

• There is Stirling's approximation the the factorial: $n! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ – Henry Dec 2 '17 at 21:10
• Or you could guess that $\frac{k!}{(k-n)!} \approx \left(k-\frac{n}2\right)^n$ – Henry Dec 2 '17 at 21:16
• @Henry that second formula looks promising, I have been playing around with it for the last half hour but cant seem to get it to where I need it to go. Do you have any pointers? – XRBtoTheMOON Dec 2 '17 at 21:58
• Then try $\frac{k!}{k^n (k-n)!} \approx \left(1-\frac{n}{2k}\right)^n = \left(1+\frac{n^2/(2k)}{n}\right)^n$ and apply the only formula you know with a suitable $x$ – Henry Dec 2 '17 at 23:54
• May be this can be helpful: $Lim_{k→∞}(1+\frac{n}{k})^k=e^{\frac{n}{k}}$ AND: $∑ ^k_{n=1}(1+\frac{n}{k})^k={\frac{k!}{n!(k-n)!}}\frac{n^n}{k^n}={\frac{k!}{k^n(k-n)!}}\frac{n^n}{n!}$ NOW expansion of n! and finding the limit of $n^n/n!$ may help – sirous Dec 3 '17 at 15:33