Distinguishable, indistinguishable paramagnetic ideal gas In the canonical ensemble, the partition function for an ideal gas is given by:
$$\frac{Z}{N!}$$
The factor $N!$ accounts for the indistinguishability of the particles of the ideal gas.
What happens if you consider a system of paramagnetic ideal gas particles, such that $N = N_\uparrow + N_\downarrow$? What does the required factor become?
I think it is:
$$\frac{Z}{N!} \frac{N!}{N_\uparrow!N_\downarrow} = \frac{Z}{N_\uparrow!N_\downarrow} $$
Can someone tell me if this is correct, and possibly explain why this factor is right in this case?
 A: Use the definition of partition function as a sum over all states. Then if you interchange two indistinguishable particles you get the same state. Now you have a standard ball into bins combinatorial problem.
A: That depends on how you define the partition function; specifically, how much you over-counted when computing it. If you were to compute the partition function using only physically distinguishable states then the $N!$ would not be necessary at all.
So, when you go to the case of spins, just how are you computing the partition function?
A: Why do you say for N particles you have N!  distinguishable state?  Lets just make sure you got it right.
Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N.
Now, if you considere that the particles are indistinguishable, for each indistinguishable states, correspond N! distinguishable states (the number of permutations). So for each indistinguishable states, you overcounted N! state.
Conclusion, you need to DIVIDE by N!.
Now if the spins can be up or down, using the same reasoning, for each indistinguishable state, correspond N1!N2! distinguishable states. Which is the number of permutation among N1 and N2. So you divide again by this new factor
As Nanite pointed, in the notation you choosed, Z is computed using distinguishability. Its not always the case, I am more used to see the N! inside the partition function Z
Btw you forgot a factorial sign 
A: The correct answer is
$$Z=\frac{z_{1}^{N}}{N!}$$
The particle themselves are indistinguishable. The spin part is just another quantum number to be accounted for in the calculation of the single particle partition function
$$z_{1}=z_{\rm space}z_{\rm spin}$$ 
