The Boolean expression $$Y = \overline{A}BC + \overline{A}C$$ can be realised with which of the following gate configurations.
A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate,
B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
C. 3-input OR gate, 3 NOT gates and one 2-input AND gate
Choose the correct answer:

Solution :

Given Boolean expression :

$$Y = \bar A BC + \bar AC$$

Taking $$\bar A C$$ common :

$$Y = \bar AC(B+1)$$

Using Boolean identity :

$$B+1 = 1$$

Therefore,

$$Y = \bar AC$$

Now check the options.

To realise :

$$Y = \bar AC$$

we need :

One NOT gate for $$A$$

One 2-input AND gate for $$\bar A$$ and $$C$$

Option A contains unnecessary gates but can realise the expression.

Option B also contains required configuration and can realise the expression.

Option C uses OR gate instead of AND gate, hence cannot realise the expression.

Final Answer :

$$A \text{ and } B$$