1. Given Boolean expression

We need to realise $Y = A\bar{B}C + \bar{A}\bar{C}$

This is a sum of two terms:

• $A\bar{B}C$
• $\bar{A}\bar{C}$

We now test each configuration.

2. Configuration A

Configuration A states:

• one 3-input AND gate
• 3 NOT gates
• one 2-input OR gate
• one 2-input AND gate

Step 2.1: Realising $A\bar{B}C$

Use:

• NOT on $B$ to get $\bar{B}$
• 3-input AND on $A, \bar{B}, C$

This gives: $A\bar{B}C$

Step 2.2: Realising $\bar{A}\bar{C}$

Use:

• NOT on $A$ to get $\bar{A}$
• NOT on $C$ to get $\bar{C}$
• 2-input AND gate on $\bar{A}, \bar{C}$

This gives: $\bar{A}\bar{C}$

Step 2.3: Add the two terms

Use the 2-input OR gate: $Y = A\bar{B}C + \bar{A}\bar{C}$

So, A is possible.

3. Configuration B

Configuration B states:

• one 3-input AND gate
• 1 NOT gate
• one 2-input NOR gate
• one 2-input OR gate

We use Boolean manipulation.

Given: $Y = A\bar{B}C + \bar{A}\bar{C}$

Notice that by De Morgan's law, $\bar{A}\bar{C} = \overline{A+C}$

A 2-input NOR gate directly gives: $\overline{A+C} = \bar{A}\bar{C}$

Also, for the first term $A\bar{B}C$ we only need one inversion, namely $\bar{B}$, so one NOT gate is sufficient.

Step 3.1: First term

• NOT on $B$ gives $\bar{B}$
• 3-input AND on $A, \bar{B}, C$ gives $A\bar{B}C$

Step 3.2: Second term

• 2-input NOR on $A$ and $C$ gives $\overline{A+C} = \bar{A}\bar{C}$

Step 3.3: OR the two terms

Using 2-input OR: $Y = A\bar{B}C + \bar{A}\bar{C}$

So, B is also possible.

4. Configuration C

Configuration C states:

• 3-input OR gate
• 3 NOT gates
• one 2-input AND gate

We check whether this can realise the expression.

Use De Morgan's law on the whole expression: $Y = A\bar{B}C + \bar{A}\bar{C}$

Then $\bar{Y} = \overline{A\bar{B}C + \bar{A}\bar{C}}$ $\bar{Y} = \overline{A\bar{B}C}\; \cdot \; \overline{\bar{A}\bar{C}}$

Now, $\overline{A\bar{B}C} = \bar{A} + B + \bar{C}$

and $\overline{\bar{A}\bar{C}} = A + C$

So, $\bar{Y} = (\bar{A}+B+\bar{C})(A+C)$

Hence, $Y = \overline{(\bar{A}+B+\bar{C})(A+C)}$

This means:

• a 3-input OR can generate $(\bar{A}+B+\bar{C})$ if we invert $A$ and $C$
• a 2-input OR can generate $(A+C)$, but configuration C has no 2-input OR, only a 2-input AND

So this direct form is not possible.

Let us check from the listed gates themselves:

• with 3 NOT gates, we can generate complements
• with one 3-input OR gate, we can generate only one sum term
• with one 2-input AND gate, we can multiply only two signals

But the given expression is an OR of two product terms, one of which is 3-input product and the other 2-input product. Without an output inversion or an extra OR gate, this set cannot realise the function directly.

Therefore, C is not possible.

5. Final selection

Possible configurations are:

• A: possible
• B: possible
• C: not possible

Hence the correct choice is: $\boxed{\text{A, B Only}}$

This corresponds to Option C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So, the answer agrees with the stored answer.