= Solution
For occupations $W_n=Wp(n)$ of $W$ distinguishable copies, the number of arrangements is the <multinomial coefficient> $\mathcal N=W!/\prod_nW_n!$. The ensemble entropy is $S_W=k_B\log\mathcal N$. <Stirling's formula> and $\sum W_n=W$ give
$$
S_W\sim k_B\left[W\log W-\sum_nWp(n)\log(Wp(n))\right]
=-Wk_B\sum_np(n)\log p(n).
$$
Dividing by the number of independent copies and taking the large-ensemble limit recovers \b[$S=-k_B\sum_np(n)\log p(n)$]. Terms with zero occupation are interpreted by the usual $0\log0=0$ convention.
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