JEE Main 2026 — 21 January, Morning Shift — Question 3

For a set with $n$ elements, the number of relations that are both reflexive and symmetric is $2^{\frac{n(n-1)}{2}}$. Here, $n = 4$, so $\frac{4 \times 3}{2} = 6$. Thus, the number of such relations is $2^{6} = 64$. Hence, the correct option is D.