= Solution
For occupation numbers zero or one, the energy and particle number are additive over modes. The <grand canonical partition function> factorizes:
$$
\mathcal Z=\prod_r\left(1+e^{-(\epsilon_r-\mu)/(kT)}\right).
$$
Dividing the product of Boltzmann weights by this product gives
$$
\boxed{\overline\rho_i=\prod_r
\frac{e^{-n_r(\epsilon_r-\mu)/(kT)}}
{1+e^{-(\epsilon_r-\mu)/(kT)}}.}
$$
The marginal mean occupation is therefore
$$
\boxed{\overline n_r=\frac1{e^{(\epsilon_r-\mu)/(kT)}+1}.}
$$
This is the <Fermi-Dirac distribution> of a noninteracting <Fermi gas>: the exclusion principle allows only zero or one fermion in each single-particle state, including its spin label.
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