Addressing inputs with select lines: In a standard multiplexer, how many select lines are required to uniquely address four (4) data input lines?

Introduction / Context:
Multiplexers use select lines to choose which one of several inputs is connected to the output. Understanding the relationship between the number of inputs and the number of select lines is a basic but essential skill in combinational logic design.

Given Data / Assumptions:

• Number of inputs N = 4.
• Binary addressing via select lines.
• Standard MUX behavior, independent of the specific logic family.

Concept / Approach:
The number of select lines k needed to address N inputs satisfies N = 2^k. Solving for k gives k = log2(N). Thus, for N = 4, k = log2(4) = 2. Therefore, two select lines (commonly labeled S1 and S0) are sufficient and necessary to address four inputs.

Step-by-Step Solution:

Verification / Alternative check:
Examine a 74HC153 (dual 4-to-1 MUX): two select lines select among four inputs for each section, confirming the theoretical relationship.

Why Other Options Are Wrong:

• Three select lines: 3 lines give 2^3 = 8 addresses, excessive for 4 inputs.
• Depends only on logic family: The relation N = 2^k is universal.
• Needs four select lines due to enable: Enable pins do not count as select lines for addressing inputs.
• “Incorrect (three select lines are not required)”: while true as a comment, it does not answer with the needed count; the correct quantitative choice is two lines.

Common Pitfalls:
Confusing enable/strobe pins with select lines; assuming one select per input instead of binary addressing.

Final Answer:
Requires two select lines (S1, S0)