Number of microstates in a two level system

Question

This sounds like a physics question but I'm sure of all the physics aspects. I'm going wrong somewhere though, in the maths. At high temperatures, the occupancy of energy levels becomes equal, so in this case both levels will have $\frac{N}{2}$ particles. The number of ways of getting $\frac{N}{2}$ particles per level is where I'm having trouble. It should be given by $$\Omega = \frac{N!}{(\frac{N}{2})!(\frac{N}{2})!} = \frac{4}{N!}$$ That line must be wrong but I have no idea why. Because what I'm aiming for is $$S = k\ln(\Omega) = Nkln(2)$$ Which my $\Omega$ doesn't give. So where have I gone wrong in finding the number of ways of getting $\frac{N}{2}$ particles per level?

Answer 1

The first part of your line, namely $$ \Omega = \frac{N!}{\frac N2!\frac N2!} $$ is correct, but $\frac N2! \ne \frac{N!}2$, which you seem to assume. You can rewrite $$ {N \choose \frac N2} = 2^N \frac{1 \cdot 3 \cdots (N-1)}{2 \cdot 4 \cdots N} $$ or approximate using Stirling $$ {N \choose \frac N2}\sim \frac{4^{\frac N2}}{\sqrt{\pi \frac N2}} $$ which gives $$ \log \Omega \approx \frac N2 \log 4 \cdot \left(-\frac 12\right) \bigl(\log \pi + \log N - \log 2\bigr) $$