Difficulty: Easy
Correct Answer: 16
Explanation:
Introduction: Digital logic gates respond to all possible binary input patterns. For a gate with n inputs, every input can be 0 or 1, so the total number of distinct input combinations equals the count of all 0/1 sequences of length n. This question checks your grasp of that fundamental counting idea for a 4-input gate.Given Data / Assumptions:
Concept / Approach: The number of binary strings of length n is 2^n because each position has 2 choices (0 or 1) and choices multiply across positions. This applies regardless of the specific gate type (AND, OR, XOR, etc.); the gate type affects outputs, not the count of input patterns.Step-by-Step Solution:
Step 1: Each input has 2 states: 0 or 1.Step 2: With n independent inputs, total combinations = 2^n.Step 3: Substitute n = 4 to get 2^4 = 16.Verification / Alternative check:
List examples mentally: 0000, 0001, 0010, ..., 1111. Counting from 0 to 15 in binary yields 16 patterns in total.Why Other Options Are Wrong:
8: This is 2^3, which would apply to a 3-input gate, not 4.32: This is 2^5, which corresponds to 5 inputs.4: This equals n, not 2^n; it ignores binary branching per input.Common Pitfalls:
Confusing number of inputs (n) with number of combinations (2^n).Assuming gate type changes the count; it does not.Final Answer:
16
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