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].

Let the inputs to the $2:4$ active-high decoder $Q$ be $A$ and $B$. Its outputs are:

• $D_0 = \overline{A}\,\overline{B}$

• $D_1 = \overline{A}B$

• $D_2 = A\overline{B}$

• $D_3 = AB$

The $4$-to-$1$ multiplexer $M$ has select lines $S_1$ (MSB) and $S_0$ (LSB), and data inputs:

• $I_0 = D_0 = \overline{A}\,\overline{B}$

• $I_1 = 0$

• $I_2 = 1$

• $I_3 = D_3 = AB$

The Boolean expression for output $Y$ is:

Now we evaluate the number of combinations of $[A, B, S_1, S_0]$ that make $Y = 1$:

1. Case $S_1S_0 = 00$: $Y = \overline{A}\,\overline{B} = 1 \implies (A, B) = (0, 0)$. This gives 1 combination: [0 0 0 0].

2. Case $S_1S_0 = 01$: $Y = 0$. This gives 0 combinations.

3. Case $S_1S_0 = 10$: $Y = 1$ regardless of $A$ and $B$. This gives all 4 combinations: [0 0 1 0], [0 1 1 0], [1 0 1 0], [1 1 1 0].

4. Case $S_1S_0 = 11$: $Y = AB = 1 \implies (A, B) = (1, 1)$. This gives 1 combination: [1 1 1 1].

Total number of combinations producing $Y = 1$ is $1 + 0 + 4 + 1 = 6$.