Spin 3/2 Statistical Mechanics Problem

I am trying to solve a problem from the book 'Introductory Statistical Mechanics' (Bowley, Sanchez). The question reads:

Calculate the free energy of a system of N particles, each with spin 3/2 with one particle per site, given that the levels associated with the four spin states have energies e, 2e, -e, -2e....

What I want to know is how I use the face that each particle has a spin 3/2? Does this add some kind of degeneracy I need to take into account?

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If there are no term in the Hamiltonian lifting the degeneracy on the spin, then you'll have a degree of degeneracy equal to $$g_s=(2s+1)$$ where s is the spin ($=\frac32$ so $g_s=4$ in your case).

Each correspond to a projection of the spin along some axis (the z usually) $$s_z = \frac32,\frac12,-\frac12,-\frac32$$

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The question actually reads: Calculate the free energy of a system of N particles, each with spin 3/2 with one particle per site, given that the levels associated with the four spin states have energies $$\frac32\epsilon,\frac12\epsilon,-\frac12\epsilon,-\frac32\epsilon$$, with degeneracies 1, 3, 3, and 1 respectively. I have tried calculating the partition as follows but this doesn't lead me to the answer.. $$Z=\exp\left(-\frac{3\epsilon}{2k_{B}T}\right)+\exp\left(\frac{3\epsilon}{2k_{B‌​}T}\right)+3\exp\left(-\frac{\epsilon}{2k_{B}T}\right)+3\exp\left(-\frac{\epsilon‌​}{2k_{B}T}\right)$$ –  Nathan Moynihan Nov 7 '12 at 12:18
(I should note that this is just for a single particle, the actual function would be to the N power) Where am I going wrong? Thanks :) –  Nathan Moynihan Nov 7 '12 at 12:20
Actually I figured it out, thanks for your help! –  Nathan Moynihan Nov 7 '12 at 15:02
Here the indication of the particle's spin $s=\frac{3}{2}$ indicates the dimension of the Hilbert space corresponding to its internal states, equal to $2s+1=4$. The energies being given imply that there is no degeneracy and that the energies are $m \epsilon$ for $m=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$. If you're told that there is in fact a degeneracy then you need to put it in by hand, but the degeneracy scheme you mention in the comments (1, 3, 3, 1) sounds more like three spin-1/2 particles being put together than one spin-3/2 particle.
To get the partition function, then you sum the factors $e^{-\beta E}$, with the correct energies: $$\zeta=\sum_{m=-3/2}^{3/2}g_m e^{-\beta \epsilon\cdot m},$$ where the degeneracies $g_m$ should all be 1 unless you're specifically told otherwise. (Note, in the $g_m\equiv1$ case, that the sum is a geometric sum and can be done explicitly.)
Since the $N$ particles are independent (unless you're told they're not) then you need to combine them appropriately to get $$Z=\frac{\zeta^N}{N!}.$$