NOR truth table counting A 3-input NOR gate has eight possible input combinations. For how many of those combinations is the output HIGH?

Introduction / Context:Counting the number of HIGH outputs across all input combinations is a quick way to check understanding of a gate’s truth table and to verify K-map groupings or SOP/POS derivations for multi-input logic gates.

Given Data / Assumptions:

• Gate type: NOR (OR followed by inversion).
• Number of inputs: 3, thus 2^3 = 8 combinations.
• Positive logic conventions (1 = HIGH, 0 = LOW).

Concept / Approach:

NOR outputs HIGH only when the internal OR would be LOW. The OR of three inputs is LOW only for the all-zero case. Therefore, the NOR gate’s output is HIGH for exactly one input combination (A = 0, B = 0, C = 0) and LOW for all others that include at least one HIGH input.

Step-by-Step Solution:

Verification / Alternative check:

Write the Boolean expression for 3-input NOR: Y = (A + B + C)’. Evaluate it at (0,0,0) → (0)’ = 1; at any other row, A + B + C = 1 → (1)’ = 0. Count of HIGH outputs is therefore 1.

Why Other Options Are Wrong:

• 2, 7, 8, or 0 do not match the NOR truth table; 7 would correspond to an AND gate’s LOW counts, not NOR’s HIGH count.

Common Pitfalls:

• Accidentally using the OR truth table (which is HIGH for 7 of 8 rows) instead of NOR.

Final Answer:

1