\(A \cap B\) = no I and no F → all M: only 1 configuration,
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- Each of these sets corresponds to assignments using only 2 types: \(2^9 = 512\). So:
- |A| = |B| = |C| = 512
- Now, intersections:
- Similarly, \(A \cap C = 1\), \(B \cap C = 1\),
- \(A \cap B \cap C = 0\) (impossible to have no types).
- By inclusion-exclusion, the number of invalid configurations (missing at least one type) is:
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