Truth-table reasoning: For a 3-input AND gate, there are 8 possible input combinations. How many of those combinations will produce a HIGH output?

Introduction / Context:AND gates output HIGH only when all inputs are HIGH. Counting how many input combinations yield a HIGH output reinforces intuition about multi-input logic behavior and probabilities in random logic testing.

Given Data / Assumptions:

• Three inputs → 2^3 = 8 combinations.
• AND function requires all inputs to be 1 for a 1 output.

Concept / Approach:Enumerate conceptually: only the combination A = 1, B = 1, C = 1 satisfies the AND condition. Every other combination contains at least one 0 and therefore forces the output to 0.

Step-by-Step Solution:Total combos = 2^3 = 8.High output cases for AND = number of “all ones” cases = 1.Thus, answer = 1.

Verification / Alternative check:Create a quick truth table and verify only the last row (1,1,1) yields HIGH.

Why Other Options Are Wrong:2 or 4: would require additional satisfying combinations which do not exist for AND.7 or 8: these are typical for OR/NOR/always-on behaviors, not AND.

Common Pitfalls:Confusing AND with OR, where many more combinations produce HIGH.

Final Answer:1