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(C) Let the output of the first $NAND$ gate be $C = \overline{A \cdot B}$.The top $NAND$ gate receives inputs $A$ and $C$,so its output is $Y_1 = \ove

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$c$ and $d$

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(C) Let the output of the first $NAND$ gate be $C = \overline{A \cdot B}$.The top $NAND$ gate receives inputs $A$ and $C$,so its output is $Y_1 = \overline{A \cdot C} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A} + (A \cdot B) = \overline{A} + B$.The bottom $NAND$ gate receives inputs $B$ and $C$,so its output is $Y_2 = \overline{B \cdot C} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B} + (A \cdot B) = \overline{B} + A$.The final output $Y$ is the $AND$ of $Y_1$ and $Y_2$,so $Y = Y_1 \cdot Y_2 = (\overline{A} + B) \cdot (\overline{B} + A)$.Evaluating for each case:- If $A=0, B=1$: $Y = (1+1) \cdot (0+0) = 1 \cdot 0 = 0$.- If $A=1, B=0$: $Y = (0+0) \cdot (1+1) = 0 \cdot 1 = 0$.- If $A=0, B=0$: $Y = (1+0) \cdot (1+0) = 1 \cdot 1 = 1$.- If $A=1, B=1$: $Y = (0+1) \cdot (0+1) = 1 \cdot 1 = 1$.Thus,$Y=1$ for combinations $c$ $(A=0, B=0)$ and $d$ $(A=1, B=1)$.

For the given logic circuit,to have the output $Y=1$,the possible combinations of inputs are:
$a) A=0, B=1$
$b) A=1, B=0$
$c) A=0, B=0$
$d) A=1, B=1$

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For the given logic circuit,to have the output $Y=1$,the possible combinations of inputs are:$a) A=0, B=1$$b) A=1, B=0$$c) A=0, B=0$$d) A=1, B=1$