Question:medium

Consider the digital circuit shown below with two input lines A and B, two select
lines S0 and S1, and an output line Y. The blocks Q and M represent active high
2:4 decoder and 4-to-1 multiplexer, respectively. Out of 16 possible input
combinations, the number of combinations that produce Y=1 is ____________.
(answer in integer)
Note: One input combination is an instance of [A B S1 S0].

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Trace which decoder output line feeds each multiplexer data input, note any branched wire that ties one decoder output into two adjacent multiplexer inputs, then for every select value S1S0 count how many (A, B) pairs make the corresponding data input HIGH.
Updated On: Aug 3, 2026
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Correct Answer: 6

Solution and Explanation

An alternative way to solve this is to build the full 4-variable truth table for \(A, B, S_1, S_0\) and evaluate \(Y\) from the decoder-to-multiplexer wiring shown in the figure, counting data-line-wise instead of row-by-row.

The 2:4 decoder produces a one-hot output based on \((A,B)\): \(D_0\) for \(00\), \(D_1\) for \(01\), \(D_2\) for \(10\), \(D_3\) for \(11\). From the diagram, the multiplexer data lines are tied as \(I_0=D_0\), \(I_1=D_1+D_2\), \(I_2=D_2+D_3\), \(I_3=D_3\), and \(Y=I_{S_1S_0}\).

Step 1: List the 4 values of \((A,B)\) against the decoder line each one activates.

AB = 00D0 active
AB = 01D1 active
AB = 10D2 active
AB = 11D3 active

Step 2: For each of the 4 values of \((S_1,S_0)\), find how many \((A,B)\) rows make \(Y=1\).

\(S_1S_0=00\) selects \(I_0=D_0\): only \(AB=00\) works, giving 1 row.

\(S_1S_0=01\) selects \(I_1=D_1+D_2\): both \(AB=01\) and \(AB=10\) work, giving 2 rows.

\(S_1S_0=10\) selects \(I_2=D_2+D_3\): both \(AB=10\) and \(AB=11\) work, giving 2 rows.

\(S_1S_0=11\) selects \(I_3=D_3\): only \(AB=11\) works, giving 1 row.

Step 3: Sum the rows across the full \(4\times4=16\)-row table: \(1+2+2+1=6\) rows have \(Y=1\).

This route, which counts how many address combinations activate each multiplexer input rather than checking each of the 16 combinations one by one, again confirms that \(Y=1\) for exactly 6 out of 16 combinations.

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