Question

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

Answer 1

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(x, y)`:

`(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(x, y) approx (x^x)/((x-y)^(x-y)) = (x/(x-y))^x (x-y)^y approx (x/x)^x (x-y)^y`
`P(x, y) approx x^y`

Returning to the original question,
`(P(N', n))/(P(N, n)) approx ((N')^n)/((N)^n) = ((N')/N)^n`

completing the derivation.