= Solution
Define $q=N_+-N_-$, where $N_+$ counts particles in the $+\epsilon$ level. Then $N_\pm=(N\pm q)/2$, so
$$
\boxed{U=\epsilon q,\qquad
S=k_B\ln\frac{N!}{[(N+q)/2]!\,[(N-q)/2]!}.}
$$
The <binomial entropy of independent two-level particles> follows from counting which of the $N$ independently labelled particles occupy each nondegenerate level. This is the usual distinguishable, or localized, two-level model; occupation constraints of a degenerate quantum gas would require different counting.
For large occupations, <Stirling's approximation> with $x=q/N$ gives
$$
\boxed{S\simeq Nk_B\left[\ln2-\frac{1+x}{2}\ln(1+x)-\frac{1-x}{2}\ln(1-x)\right].}
$$
Allowed $q$ runs from $-N$ to $N$ in steps of two. <Entropy> is largest near equal populations and is zero when all particles occupy one level; use $0\ln0=0$ at the endpoints.
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