Canonical partition sum for two fermions in harmonic potential In an old exam, I found the following problem:

Two Particles in a potential well
We look at a onedimensional harmonic potential well that hold two spinless particles that do not interact with each other. Each particle has the energy eigenvalues $E_n = \hbar\omega(n + 1/2)$ with $n = 0, 1, \ldots$.
  
  
*
  
*The particles are distinguishable. Show that the canonical partition sum $Z_\text C$ factorizes, then calculate $Z_\text C$.
  
*The two particles are now indistinguishable fermions. Calculate the canonical partition sum $Z_\text C$ for this case.

I got the first one done, that was pretty straightforward:
My derivation http://chaos.stw-bonn.de/users/mu/uploads/2014-02-12/Bildschirmfoto3.png
For the second part, I am not sure what my states are now. Since the particles are fermions, they will have Spin $s = 1/2$. That means that the wavefunction describing both particles will have to be totally antisymmetric regarding particle exchange.
If they both occupy the same $n$ and their spins are opposing, this should be fine. What other states are allowed? If the spins are in the same direction, $n_1 \neq n_2$ for sure. Since the spin part of the wavefunction is symmetric, the other part has to be antisymmetric. The only other part that I could think of would be the spatial part of the wavefunction. If $n_1$ and $n_2$ are odd and even, respectively, each spatial wavefunction would change its sign when the particles are exchanged. However, since $n_1$ and $n_2$ are different, the spatial wavefunctions would look different as well, having a different amount of zeros (i. e. $|\psi|^2 = 0$).
Which states are allowed?
 A: First, note that the following statement is untrue:

Since the particles are fermions, they will have spin $s=1/2$.

Fermions can have any spin in the set $\{\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots\}$, namely in the set of non-negative integers plus a half.
Second, my interpretation of the problem is that the author wants you to ignore the spin part of the states of the particles.  I think this interpretation is given credence by the fact that the spin of the fermions is left unspecified, and the question would therefore be ambiguous if the intention were for you to include the spin factor in the Hilbert space.
If we ignore spin, then the enumeration of the states can be accomplished as follows. The full Hilbert space $\mathcal H$ for distinguishable particles is spanned by (tensor) product basis elements $|n_1\rangle|n_2\rangle$.  So given any $|\psi\rangle$ of the system of distinguishable particles, there exist complex numbers $c_{n_1, n_2}$ for which
\begin{align}
  |\psi\rangle = \sum_{n_1, n_2}c_{n_1,n_2}|n_1\rangle|n_2\rangle
\end{align}
The Hilbert space $\mathcal H^-$ for identical Fermions (with spin ignored) is the subspace of this Hilbert space consisting of states $|\psi\rangle$ which satisfy $P|\psi\rangle = -|\psi\rangle$ where $P$ is the exchange operator defined as the unique linear operator whose action on the product basis elements is $P|n_1\rangle |n_2\rangle = |n_2\rangle|n_1\rangle$.  If we apply this to $|\psi\rangle$, then we find that
\begin{align}
  c_{n_1,n_2} = -c_{n_2,n_1}
\end{align}
and this tells us that $c_{n,n} = 0$ for every $n\geq 0$.  These conditions allow us to write the general such state as
\begin{align}
  |\psi\rangle = \sum_{n_1\neq n_2} c_{n_1,n_2}\Big(|n_1\rangle|n_2\rangle - |n_2\rangle|n_2\rangle\Big)
\end{align}
This gives the answer to your question

Which states are allowed?

A basis of states for the fermionic subspace is
\begin{align}
  \Big\{|n_1\rangle|n_2\rangle - |n_2\rangle|n_1\rangle \, \Big|\, n_1\neq n_2\Big\}
\end{align}
The first few of these guys (in order of non-decreasing energy from left to right) is
\begin{align}
  |1\rangle|0\rangle - |0\rangle|1\rangle, \qquad
  |2\rangle|0\rangle - |0\rangle|2\rangle , \qquad
  |3\rangle|0\rangle - |0\rangle|3\rangle, \qquad
  |1\rangle|2\rangle - |2\rangle|1\rangle.
\end{align}
There's actually a really cool thing you can do though.  There's a notation for these states that is a lot more intuitive.  Let one of these states $|n_1\rangle|n_2\rangle-|n_2\rangle|n_1\rangle$ be given, then we can write this as
\begin{align}
  |0,\dots 0, 1, 0\dots, 0,1,0\dots\rangle
\end{align}
where there is a $1$ in spot $n_1$ in the sequence and a $1$ in spot $n_2$ of the sequence.  For example, the state $|0\rangle|2\rangle-|2\rangle|1\rangle$ would be written
\begin{align}
  |1,0,1,0,\dots\rangle
\end{align}
and the state $|1\rangle|5\rangle-|1\rangle|5\rangle$ would be written
\begin{align}
  |0,1,0,0,0,1,0,\dots\rangle
\end{align}
So the basis for the fermionic subspace $\mathcal H^-$ can also be identified with the set of sequences of $0$'s and $1$'s, where the sum of the sequence is $2$.  Intuitively, you can think of each element of the sequence as one of the energy states of the potential, and you can think of placing one of each of the particles in one of these levels, and that's represented by a $1$ in the sequence, while the unfilled levels are assigned $0$'s.  Since there are $2$ particles, you will have two $1$'s in the sequence, and the rest zeros.  The fact that you can only have $1$'s and $0$'s comes from the fact that the system consists of fermions, so antisymmetry tells us that they can't be in the same quantum state at the same time.  In other words, it's just the Pauli Exclusion Principle.
