Discussion
For 3 : 8 decoder, the LED will be switched on for :
Answer: C
The LED is on for \(\bar{C}\bar{B}A + \bar{C}BA + C\bar{B}\bar{A} + CB\bar{A}\).
The decoder outputs are active-low, and y0, y2, y5 and y7 feed the NAND gate. A NAND of active-low signals acts as an OR of the selected minterms, so
\[ F = m_{0} + m_{2} + m_{5} + m_{7} \]
The LED has its anode at +VCC and its cathode at F, so it lights only when F is LOW, that is for the remaining minterms m1, m3, m4 and m6. With C as the most significant bit these are 001, 011, 100 and 110, the four terms of the third option.
The other options include terms such as CBA (m7) or \(\bar{C}B\bar{A}\) (m2), for which F is HIGH and the LED is off.
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