= Solution
Use the ideal <Langmuir adsorption model>: identical independent sites, at most one atom per site, no lateral interactions, and no additional thermally active adsorbed degrees of freedom beyond the two <spin> states. Treat the atoms as indistinguishable and localized after <adsorption>; only site occupancy and <spin> choices distinguish configurations.
There are $\binom MN$ ways to select $N$ occupied sites. For any such selection, independent <spin> sums contribute $z^N$. Thus
$$
\boxed{Q_N=\binom MN z^N.}
$$
With $q_{\rm ads}=Mz$ as defined in part (b), this is
$$
\boxed{Q_N=q_{\rm ads}^{\,N}f(M,N),\qquad
f(M,N)=\frac{M!}{N!(M-N)!\,M^N},\quad0\leq N\leq M.}
$$
Equivalently, starting from labelled particles gives $M!/(M-N)!$ distinct assignments to different sites; division by $N!$ removes particle relabellings. The $M^{-N}$ removes the site multiplicity already included in each full one-particle function.
If $q_{\rm ads}$ denotes the one-site internal factor instead, \b[$f(M,N)=\binom MN$] and $Q_N=\binom MN q_{\rm ads}^{\,N}$. These are identical physical results, not different counting models. In particular, the complete-function convention gives $f(M,1)=1$ and hence $Q_1=q_{\rm ads}$.
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