Occupied lattice sites, determining number of microstates and energy

Question

A solid consisting of $N$ molecules on a lattice of $N$ sites is isolated from its environment, and has energy $E$. Each molecule is fixed in position and independent of all others. It can be in any one of four internal states. Two of these states have energy $0$, while the other two states have energy $b$.

Consider $P=E/b$, so that $P$ is the number of molecules with energy $b$.

(a) Calculate the number of microstates of the solid when it has energy $E$.

(b) The solid is now placed in good thermal contact with its environment at temperature $T$. What is the solid's mean energy?

(c) Suppose $b=0.1 \mathrm{eV}$. What is the energy per molecule at $25^{\circ}C$?

I believe the number of microstates of the solid when it has energy $E$ is ${N/2 \choose bP}$.

However, I am not sure. Can someone clarify/guide, and assist with the latter parts?

Answer 1

Part B)

Let's find the partition function.

$Z = \Sigma_i{e^{-E_i/Tk_b}} $ over all microstates i.

We have 4 microstates of energy b,b,0 and 0. Therefore:

$Z = 2e^{-b/Tk_b} + 2 = 2e^{-b\beta} + 2$

Now we need to find the energy. To do this we use the formula:

$E = -\frac{d}{d\beta}Z(\beta)$ = $-\frac{d}{d\beta}(2e^{-b\beta} + 2)$

Evaluating gives:

$<E> = 2be^{-b\beta} = 2be^{b/k_bT}$

Part C)

Substitute the numbers given into the above expression for E. This expression is already "per particle"