JEE Mathematics 2026 Question with Solution

Given: The set is $A = \{a, b, c, d\}$ so $|A| = 4$.

Find: The number of relations on $A$ that are both reflexive and symmetric.

A relation on $A$ is a subset of $A \times A$.

For a reflexive relation, all diagonal pairs must be included:

$$(a,a), (b,b), (c,c), (d,d)$$

So these $4$ pairs are fixed, giving only $1$ choice for each.

Now consider the off-diagonal pairs. Since $|A \times A| = 4^2 = 16$, the number of off-diagonal elements is

$$16 - 4 = 12$$

Because the relation must be symmetric, pairs of the form $((x,y),(y,x))$ must be chosen together. Thus the $12$ off-diagonal ordered pairs form

$$\frac{12}{2} = 6$$

symmetric pair-groups.

For each such group, there are exactly $2$ choices:

1. include both,
2. include neither.

Hence total number of such relations is

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

In general, for a set of size $n$, the count is

$$2^{\frac{n(n-1)}{2}}$$

Therefore, the total number of relations is $64$ and the correct option is B.

Given: A set with $4$ elements.

Find: How many relations are simultaneously reflexive and symmetric.

Reflexivity forces all pairs $(x,x)$ to be present, so there is no freedom on the diagonal.

For distinct elements, the possible unordered pairs are:

$$\{a,b\}, \{a,c\}, \{a,d\}, \{b,c\}, \{b,d\}, \{c,d\}$$

There are $\binom{4}{2} = 6$ such choices.

Each unordered pair corresponds to the ordered pair set

$$\{(x,y),(y,x)\}$$

For symmetry, either both are included or both are excluded. So each unordered pair contributes $2$ choices.

Therefore,

$$2^6 = 64$$

Hence the correct option is B.