0 0 votes The minimum number of $2$-input $\text{NAND}$ gates required to implement the Boolean function $Z=A \bar{B} C$, assiming that $A, B$ and $C$ are available, is two three five six Combinational Circuits gate1998-ec combinational-circuits logic-gates numerical-answers + – admin 337 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.