<s>[INST] Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $ I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$  The result doesn't seem right!
Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.
$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$
I would appreciate your input!
 [/INST] The easiest way to get the correct approximation is to use Stirling in the logarithmic form:
$$\ln n!=n(\ln n-1)+\frac12 \ln \left(2\pi n\right)+\frac{1}{12n}+O(n^{-3}),$$
which also implies
$$\ln\frac{n!}{(n-2x)!}=2x \ln n+\frac{x-2x^2}{n}+O(n^{-2}).$$
Now using that
$$\ln (n-1)^{-2x}=-2x\ln(n-1)=-2x\ln n+\frac{2x}{n}+O(n^{-2}),$$
we get
$$\ln\left[\frac{n!}{(n-2x)!}(n-1)^{-2x}\right]=\frac{3x-2x^2}{n}+O(n^{-2}).$$
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