Right choice is (b) We need two NOT gates obtained from NAND gates and one NAND gate
The explanation: We need OR gate (Y = A + B) from NAND gate (Y = not (A.B))
Y = not (not (A.B))
Y = not (not (A)) + not (not (B)) [Using the Boolean identify ➔ not (A.B) = not (A) + not (B)]
So, Y = A + B [Since not (not (A)) = A and not (not (B)) = B]
Therefore, to obtain an OR gate from a NAND gate, we need two NOT gates obtained from NAND gates and one NAND gate.