The given combination represents the following gate:

The correct option is OR.

Step-by-Step Explanation:

By analyzing the provided image, we can identify three NAND gates connected to form a combinational logic circuit. Let's analyze the behavior of this circuit step-by-step:

1. Analyzing the First (Top-Left) NAND Gate:
The inputs of this NAND gate are shorted (tied together) to a single input terminal A.
When the inputs of a NAND gate are tied together, it functions as a NOT gate.
Therefore, the output of this gate is the complement of A:

$\overline{A}$

2. Analyzing the Second (Bottom-Left) NAND Gate:
Similarly, the inputs of this NAND gate are tied together to a single input terminal B, making it act as a NOT gate.
The output of this gate is the complement of B:

$\overline{B}$

3. Analyzing the Third (Right) NAND Gate:
The outputs from the first two gates, $\overline{A}$ and $\overline{B}$, serve as the inputs to this third NAND gate.
The output of a NAND gate is the complement of the product (AND operation) of its inputs. Therefore, the final output Y is given by:

$Y=\overline{\overline{A}·\overline{B}}$

4. Simplifying the Output using De Morgan's Law:
According to De Morgan's theorem, the complement of a product is equal to the sum of the complements:

$\overline{X·Y}=\overline{X}+\overline{Y}$
Applying this to our expression where $X=\overline{A}$ and $Y=\overline{B}$:

$Y=\overline{\overline{A}}+\overline{\overline{B}}$
Since a double complement cancels out ($\overline{\overline{A}}=A$ and $\overline{\overline{B}}=B$), the expression simplifies to:

$Y=A+B$
This is the standard Boolean algebraic expression for a two-input OR gate.

5. Truth Table Verification:
We can verify this result by looking at all possible combinations of inputs:
• If $A=0$, $B=0$: then $\overline{A}=1$, $\overline{B}=1$ ⇒ $Y=\overline{1·1}=0$
• If $A=0$, $B=1$: then $\overline{A}=1$, $\overline{B}=0$ ⇒ $Y=\overline{1·0}=1$
• If $A=1$, $B=0$: then $\overline{A}=0$, $\overline{B}=1$ ⇒ $Y=\overline{0·1}=1$
• If $A=1$, $B=1$: then $\overline{A}=0$, $\overline{B}=0$ ⇒ $Y=\overline{0·0}=1$

The output matches the truth table of a standard OR logic gate.