# Show that? (P(N', n))/(P(N,n)) ~~ ((N')/N)^n (N', N, n) in ZZ^+ (N', N )">>" n P(x,y) = (x!)/((x-y)!) Stirling Approx allowed

Jul 18, 2018

Stirling's approximation says that
x! approx sqrt(2pi n) (n/e)^n = sqrt(2pi) * sqrt(n) * n^n * e^-n

In this way, we can approximate $P \left(x , y\right)$:

(x!)/((x-y)!) approx sqrt(2pi)/sqrt(2pi) cdot sqrt(x/(x-y)) cdot (x^x/(x-y)^(x-y)) e^(-x)/e^(-(x-y))

Since the second condition says that $y$ is far less than $x$ for our purposes, so
$\sqrt{x} \approx \sqrt{x - y}$
And since all of these we know that $x$ and $x - y$ are far greater than $e$, it is clear that the dominating term is
$P \left(x , y\right) \approx \frac{{x}^{x}}{{\left(x - y\right)}^{x - y}} = {\left(\frac{x}{x - y}\right)}^{x} {\left(x - y\right)}^{y} \approx {\left(\frac{x}{x}\right)}^{x} {\left(x - y\right)}^{y}$
$P \left(x , y\right) \approx {x}^{y}$

Returning to the original question,
$\frac{P \left(N ' , n\right)}{P \left(N , n\right)} \approx \frac{{\left(N '\right)}^{n}}{{\left(N\right)}^{n}} = {\left(\frac{N '}{N}\right)}^{n}$

completing the derivation.