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Boolean algebra practice: For the Boolean expression A + B·C (with variables A, B, C), how many distinct minterms (input combinations) satisfy the expression?

Difficulty: Easy

Correct Answer: 5

Explanation:

Introduction / Context:
This digital logic question checks your understanding of minterms and coverage in a three-variable Boolean expression. Knowing how to count satisfying input combinations (minterms) is foundational for Karnaugh maps, simplification, and logic synthesis.

Given Data / Assumptions:

Expression: A + BC where A, B, C are Boolean variables.
Domain: All 2^3 = 8 possible input combinations for (A, B, C).
Operator precedence: '·' means AND, '+' means OR.

Concept / Approach:
A term A in an OR expression A + X covers all cases where A = 1 (regardless of B, C). The product BC contributes additional cases when A = 0 and both B and C are 1. We must avoid double counting overlaps.

Step-by-Step Solution:

Count cases with A = 1: there are 2^2 = 4 minterms (for all combinations of B, C).Now consider A = 0. Then the expression reduces to BC. Only (B, C) = (1, 1) satisfies BC = 1, which adds exactly 1 new minterm.Total minterms = 4 (from A = 1) + 1 (from A = 0 and BC = 1) = 5.

Verification / Alternative check:
List explicitly: A = 1 with (B, C) ∈ {(0,0), (0,1), (1,0), (1,1)} → 4 minterms. Plus A = 0, B = 1, C = 1 → 1 minterm. No other A = 0 case works. Sum = 5.

Why Other Options Are Wrong:

2 or 3: undercount by ignoring A = 1 coverage or the extra BC case at A = 0.4: counts only A = 1 cases and misses the BC case at A = 0.6: overcounts (there is only one additional case when A = 0).

Common Pitfalls:

Double counting the case A = 1, B = 1, C = 1 (already counted in A = 1 set).Forgetting that BC only matters when A = 0.

Boolean algebra practice: For the Boolean expression A + B·C (with variables A, B, C), how many distinct minterms (input combinations) satisfy the expression?

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