The minimum number of NAND gates required for implementing the Boolean expression \(AB+A\, \overline{B}C+A\, \overline{B}\, \overline{C}\)
\(AB + A\bar{B}C + A\bar{B}\bar{C} = AB + A\bar{B}(C + \bar{C}) = AB + A\bar{B} = A(B + \bar{B}) = A\)
The output is just A, a direct wire, so no gates are needed.
Answer: (B) 0
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