Alternatively, each object can go into one of the two subsets, giving \(2^7 = 128\) labeled assignments, but dividing by 2 for indistinguishability gives \(64\), and subtracting the two cases where all are in one zone (leaving the other empty), we get \(64 - 1 = 63\) (we subtract 1 for the all-empty-one-case, but actually, each symmetric split is counted once, so the standard result applies directly).

Wed, 29 Apr 2026 18:02:36 +0000

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