0 votes 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 Others gate1998-ec + – admin asked Sep 26, 2022 • edited Dec 1, 2023 by makhdoom ghaya admin 46.4k points 58 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.