JEE Mathematics 2025 Question with Solution

Given: The set is $A = \{1,2,3\}$. The relation must contain $(1,1), (2,2), (3,3)$ because it is reflexive, and it also includes $(1,2)$. We need relations having at most $6$ elements that are reflexive and transitive but not symmetric.

Find: The number of such relations.

The required pairs already present are

$$(1,1), (2,2), (3,3), (1,2)$$

The remaining possible ordered pairs are

$$(2,1), (2,3), (1,3), (3,1), (3,2)$$

Now count casewise.

For exactly $4$ elements, no further pair is added. So the relation is

$$\{(1,1),(2,2),(3,3),(1,2)\}$$

This is reflexive and transitive, and it is not symmetric because $(2,1)$ is not included. Hence there is $1$ way.

For exactly $5$ elements, one more pair is added to the required four. From the solution working, the valid possibilities contribute $2$ ways.

For exactly $6$ elements, two more pairs are added to the required four. From the solution working, the valid possibilities contribute $3$ ways.

Therefore,

$$\text{Total number of ways} = 1 + 2 + 3 = 6$$

So, the required number of relations is $6$.

Given: $A = \{1,2,3\}$ and $(1,2) \in R$. Since $R$ is reflexive, we must include

$$(1,1), (2,2), (3,3)$$

So every valid relation already has the four pairs

$$(1,1), (2,2), (3,3), (1,2)$$

Find: How many such relations have at most $6$ elements, are transitive, and are not symmetric.

The other available ordered pairs are

$$(2,1), (2,3), (1,3), (3,1), (3,2)$$

Now use the count stated in the extracted solution.

1. Exactly $4$ elements: Only the compulsory four pairs are present. This gives $1$ way.

2. Exactly $5$ elements: The solution states there are $2$ ways.

3. Exactly $6$ elements: The solution states there are $3$ ways.

Hence,

$$\begin{aligned} \text{Total} &= 1 + 2 + 3 \\ &= 6 \end{aligned}$$

Therefore, the required numerical value is $6$.