JEE Mathematics 2024 Question with Solution

Given: The set is $A = \{1, 2, 3, 4\}$, so $n = 4$.

Find: The number of symmetric relations on $A$ that are not reflexive.

A relation $R$ is symmetric if $(a,b) \in R \implies (b,a) \in R$. A relation is reflexive if $(a,a) \in R$ for all $a \in A$.

For a symmetric relation on a set of size $4$:

• the $4$ diagonal pairs $(1,1), (2,2), (3,3), (4,4)$ can be chosen independently,
• the off-diagonal pairs are grouped as unordered pairs $\{(a,b),(b,a)\}$.

There are

$$\binom{4}{2} = 6$$

such off-diagonal groups.

Hence, the total number of symmetric relations is

$$2^4 \times 2^6 = 2^{10} = 1024$$

For a relation to be both symmetric and reflexive, all diagonal pairs must be included, so there is only one choice for the diagonal entries, while the $6$ off-diagonal symmetric groups can still be chosen freely:

$$1 \times 2^6 = 64$$

Therefore, the number of symmetric relations that are not reflexive is

$$1024 - 64 = 960$$

Therefore, the required number of relations is $960$.

Given: $A = \{1,2,3,4\}$.

Find: The number of relations on $A$ that are symmetric but not reflexive.

The Cartesian product $A \times A$ has

$$4^2 = 16$$

ordered pairs.

These are of two types:

1. Diagonal elements: $(1,1), (2,2), (3,3), (4,4)$, so there are $4$ of them.
2. Off-diagonal elements: pairs $(a,b)$ with $a \ne b$.

For symmetry, whenever one of $(a,b)$ and $(b,a)$ is included, the other must also be included. So the off-diagonal entries are chosen in pairs. The number of such pairs is

$$\binom{4}{2} = 6$$

Now count symmetric relations:

• each of the $4$ diagonal elements has $2$ choices: included or not included,
• each of the $6$ off-diagonal symmetric pairs has $2$ choices: include both or include neither.

Thus,

$$\text{Total symmetric relations} = 2^4 \cdot 2^6 = 2^{10} = 1024$$

Now count symmetric and reflexive relations:

• reflexive means all $4$ diagonal elements must be present, so only $1$ choice,
• the $6$ off-diagonal symmetric pairs still have $2$ choices each.

Hence,

$$\text{Symmetric and reflexive relations} = 1 \cdot 2^6 = 64$$

So the required number is

$$\text{Symmetric but not reflexive} = 1024 - 64 = 960$$

Therefore, the answer is $960$.