Partition function for classical indistinguishable particles and Bose particles We have two particles that can be in either level $E_0 = 0$ or in level $E_1$.
If we treat them as Bose particles, then the partition function will be:
$$ Z = 1 + e^{-\beta E_1} + e^{-2\beta E_1}, $$
whereas if we treat them as classical indistinguishable particles we'd get:
$$ Z = \frac{(1+e^{-\beta E_1})^2}{2!} = \frac{1}{2} + e^{-\beta E_1} + \frac{e^{-2\beta E_1}}{2}. $$
Why the discrepancy?
 A: In quantum statistics there there is never any over-counting. In this system containing two particles (= a quantum many-body system) the only three possible many-body states are
$$\vert \psi_A \rangle = \vert E_0 \rangle \otimes \vert E_0 \rangle \qquad \text{with energy} \quad E = 0$$ (with which I denote particle one being in state $E_0$ and particle two also being in state $E_0$) or $$\vert \psi_B \rangle = \frac{1}{\sqrt{2}}\left(\vert E_0 \rangle \otimes \vert E_1 \rangle + \vert E_1 \rangle \otimes \vert E_0 \rangle\right) \quad \text{(bosonic symmetry) with energy}  \quad E = E_1$$ or $$\vert \psi_C \rangle = \vert E_1 \rangle \otimes \vert E_1 \rangle \qquad \text{with energy} \quad E = 2E_1$$
(The notion of particles in a quantum many-body system is dubious.) Hence the correct solution is $$ Z = 1 + e^{-\beta E_1} + e^{-2\beta E_1}.$$
Alternatively, you could think about having two isolated quantum systems (far away from each other), in which case they are in effect distinguishable (because they are far away from each other). In this case the partition function is $$Z = \left(1 + e^{-\beta E_1}\right)^2 = 1 + 2e^{-\beta E_1} + e^{-2\beta E_1}.$$
However, in classical statistical mechanics this question is a tricky one. As far as I know, we always decide to 'forget' all information of individual particles and just treat them as an ensemble with a given temperature, pressure, etc (macroscopic observables). Writing down a partition function and doing classical statistical mechanics with distinguishable particles doesn't really make sense - hence the discrepancy.
The Gibbs factor of $N!$ is just a way of recovering the correct answers in classical thermodynamics, it does not arise from considerations of real distinguishability.
A: It comes from the first partition, your not considering all possible 2-states. Indeed, one possible state is both are in $0$, 2 other are one is in $0$ the other in $E_1$ and finally both in $E_1$. Because they are indistinguishable you must divide the whole thing by 2!.
