Official PYQ · 2026Set 2 · Q49Combinational CircuitsNAT2 marks

GATE CS 2026 Set 2 Q49 — Combinational Circuits

Consider the digital circuit shown below with two input lines $\text{A}$ and $\text{B}$ , two select lines $\text{S0}$ and $\text{S1}$ , and an output line $\text{Y}$ . The blocks $\text{Q}$ and $\text{M}$ represent active high $2:4$ decoder and $4$ -to- $1$ multiplexer, respectively. Out of $16$ possible input combinations, the number of combinations that produce $\mathrm{Y}=1$ is $\_\_\_\_\_\_$ (answer in integer)

Note: One input combination is an instance of [ $\text{A B S1 S0}$ ].

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Answer: 6:6

The active-high $2:4$ decoder $Q$ with inputs $A$ (MSB) and $B$ (LSB) produces the following outputs:

$D_0 = \bar{A}\bar{B}$
$D_1 = \bar{A}B$
$D_2 = A\bar{B}$
$D_3 = AB$

From the circuit diagram, the inputs to the $4$ -to- $1$ multiplexer $M$ are:

$I_0 = I_2 = D_1 + D_3 = \bar{A}B + AB = B$ (via the OR gate)
$I_1 = D_0 = \bar{A}\bar{B}$
$I_3 = D_2 = A\bar{B}$

The output $Y$ of the multiplexer with select lines $S_1, S_0$ is given by:

Y = \bar{S_1}\bar{S_0}I_0 + \bar{S_1}S_0I_1 + S_1\bar{S_0}I_2 + S_1S_0I_3

Substituting the values, we analyze the combinations of $[A\ B\ S_1\ S_0]$ that make $Y = 1$ for each select line state:

$S_1S_0 = 00$ : $Y = B \implies$ $2$ combinations: $(0, 1, 0, 0)$ and $(1, 1, 0, 0)$ .
$S_1S_0 = 01$ : $Y = \bar{A}\bar{B} \implies$ $1$ combination: $(0, 0, 0, 1)$ .
$S_1S_0 = 10$ : $Y = B \implies$ $2$ combinations: $(0, 1, 1, 0)$ and $(1, 1, 1, 0)$ .
$S_1S_0 = 11$ : $Y = A\bar{B} \implies$ $1$ combination: $(1, 0, 1, 1)$ .

Summing these up, the total number of combinations producing $Y = 1$ is $2 + 1 + 2 + 1 = 6$ .

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Published 10 Oct 2026

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