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(C) The output $Y$ of the given logic circuit is determined as follows:$1$. The input $A$ passes through a $NOT$ gate to become $\overline{A}$.$2$. Th

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$7$

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(C) The output $Y$ of the given logic circuit is determined as follows:$1$. The input $A$ passes through a $NOT$ gate to become $\overline{A}$.$2$. The inputs $\overline{A}$ and $B$ are fed into a $NAND$ gate,producing $\overline{\overline{A} \cdot B}$.$3$. The input $C$ passes through a $NOT$ gate to become $\overline{C}$.$4$. The outputs $\overline{\overline{A} \cdot B}$ and $\overline{C}$ are fed into a $NOR$ gate,resulting in $Y = \overline{(\overline{\overline{A} \cdot B}) + \overline{C}}$.$5$. Using De Morgan's theorem,$Y = \overline{(\overline{\overline{A} \cdot B})} \cdot \overline{(\overline{C})} = (\overline{A} \cdot B) \cdot C = \overline{A} \cdot B \cdot C$.$6$. The output $Y=1$ only when $\overline{A}=1, B=1, C=1$,which means $A=0, B=1, C=1$.$7$. There are $2^3 = 8$ total possible combinations for inputs $A, B, C$.$8$. Since $Y=1$ for only $1$ combination,the number of combinations for which $Y=0$ is $8 - 1 = 7$.

Consider the logic circuit with inputs $A, B, C$ and output $Y$. How many combinations of $A, B$ and $C$ give the output $Y=0$?

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