Truth table size for a multi-input OR gate: An OR gate has 6 independent inputs. How many distinct input rows (“input words”) appear in its truth table?

Introduction / Context:
The size of a truth table depends solely on the number of independent binary inputs, not on the specific logic function. Each binary input can be 0 or 1, so the total number of combinations grows exponentially with the number of inputs.

Given Data / Assumptions:

• Gate type: OR (but table size is independent of the function).
• Number of inputs n = 6.
• Inputs are binary (0 or 1).

Concept / Approach:
The number of distinct input combinations is 2^n. This counts every pattern (000000 through 111111 for 6 inputs). Therefore, the OR gate’s truth table has exactly 2^6 rows of input combinations, even though many of those produce the same output value (1) for OR.

Step-by-Step Solution:
Use formula: count = 2^n.Substitute n = 6 → 2^6 = 64.Thus, 64 input rows appear in the truth table.

Verification / Alternative check:
Enumerate powers of two: 2^4=16, 2^5=32, 2^6=64, 2^7=128. With six inputs, the correct count is 64, matching 2^6.

Why Other Options Are Wrong:
32 and 16 correspond to 5 and 4 inputs respectively; 128 corresponds to 7 inputs. “None of the above” is incorrect because 64 is correct.

Common Pitfalls:
Confusing output patterns with input rows; the output column content does not change the number of input combinations.

Final Answer:
64