= Solution
At sufficiently low <packing fraction>, neglect overlaps of the excluded regions around different disc centres. Sequential insertion gives the configurational <integral>
$$
\mathcal Z_N\simeq\prod_{j=0}^{N-1}(a-4ja_d).
$$
Taking logarithms and expanding to first order in density,
$$
\ln\mathcal Z_N\simeq N\ln a-\frac{4a_d}{a}\sum_{j=0}^{N-1}j
=N\ln a-\frac{2a_dN(N-1)}a.
$$
For large $N$, replace $N-1$ by $N$. The expression $N\ln(a-Nb)$ has the same leading-density expansion when \b[$b=2a_d$]. Thus the <hard-disc excluded-area approximation> yields
$$
\boxed{Q_{\rm ld}\simeq\frac{(a-2Na_d)^N}{\Lambda^{2N}N!}}.
$$
The factor $N!$ accounts for indistinguishable particles and $\Lambda^{2N}$ for two-dimensional momentum integration. The factor one-half between pair-excluded area and $b$ avoids counting every pair twice. Equality of leading expansions does not make this free-volume resummation exact at high density.
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