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Answer: 2. 1, 0

Gate P is an AND gate on A and B; gate Q is a NAND gate whose inputs are P's output and C. So \(Y=\overline{(A\cdot B)\cdot C}\).

  • A = B = C = 0: \(P=0\), \(Y=\overline{0\cdot0}=1\).
  • A = B = C = 1: \(P=1\), \(Y=\overline{1\cdot1}=0\).

Output is 1 then 0. Option (2).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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