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(B) Let the inputs $A, B$ and $C$ be given to the circuit.Gate-$I$ is an $AND$ gate,and Gate-$II$ is a $NAND$ gate.The output of the $AND$ gate is $X

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$0$,$1$

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(B) Let the inputs $A, B$ and $C$ be given to the circuit.Gate-$I$ is an $AND$ gate,and Gate-$II$ is a $NAND$ gate.The output of the $AND$ gate is $X = A \cdot B$.The final output $Y$ is the $NAND$ of $X$ and $C$,so $Y = \overline{X \cdot C} = \overline{(A \cdot B) \cdot C}$.When $A=1, B=1, C=1$ (all high):$Y = \overline{(1 \cdot 1) \cdot 1} = \overline{1 \cdot 1} = \overline{1} = 0$.When $A=0, B=0, C=0$ (all low):$Y = \overline{(0 \cdot 0) \cdot 0} = \overline{0 \cdot 0} = \overline{0} = 1$.

For the following combination of logic gates,when all the three inputs $A$,$B$ and $C$ are first 'high' and then 'low',the output $Y$ will be respectively.

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