How can I prove that, for [itex]N\gg n[/itex]
[itex]\frac{N!}{(N-n)!}\approx N^{n}[/itex]
I've tried doing
[itex]\frac{N!}{(N-n)!}=\exp\left(\ln\frac{N!}{(N-n)!}\right)=\exp\left(\ln N!-\ln\left(N-n\right)!\right)[/itex]
[itex]\underset{stirling}{\approx}\exp\left(N\ln N-N-\left(N-n\right)\ln\left(N-n\right)+N-n\right)[/itex]
[itex]=N^{N}+\left(N-n\right)^{n-N}+\exp-n[/itex]
But it doesn't look like I'm getting there
[itex]\frac{N!}{(N-n)!}\approx N^{n}[/itex]
I've tried doing
[itex]\frac{N!}{(N-n)!}=\exp\left(\ln\frac{N!}{(N-n)!}\right)=\exp\left(\ln N!-\ln\left(N-n\right)!\right)[/itex]
[itex]\underset{stirling}{\approx}\exp\left(N\ln N-N-\left(N-n\right)\ln\left(N-n\right)+N-n\right)[/itex]
[itex]=N^{N}+\left(N-n\right)^{n-N}+\exp-n[/itex]
But it doesn't look like I'm getting there
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