How to calculate the number of microstates in quantum-mechanical harmonic oscillators (microcanonical)?

Question

So I have the question where a system of two QM oscillators both have energies $\left(n+\frac12\right)\hbar\omega_0$, and the total energy of the system is given to be $E'=\left(n'+\frac12\right)\hbar\omega_0$ where both $n$ and $n'$ are positive integers. I am asked to find the number of microstates $\Omega$.

Now my first guess was that $\Omega=\frac{\left(n'+1\right)!}{n'!}=n'+1$ . I got this by taking the "stars and bars" approach.

However on the given solution it is given that $\Omega=\frac{(n')!}{\left(n'-1\right)!}=n'$.

Can anyone explain to me why my approach is wrong and the reasoning behind the approach given?

As a side note, I know that at very high energies the extra "$+1$" becomes negligible this is just to get my reasoning correct.

Answer 1

yes so the microcanonical partition function is

$\Omega = \frac{\mathrm{Number~of~Macrostates}}{\mathrm{Number~of~Microstates}}$

So the $n'!$ is your number of macrostates (which is always the case for all systems normally it is n!).

The $(n'-1)!$ comes by avoiding double counting.