JEE Main 2024 — 29 January, Shift 2 — Question 18

The set is $\{1,2,3,4\}$. The smallest equivalence relation containing $(1,2)$ and $(1,3)$ must be reflexive, symmetric, and transitive. Reflexive pairs: $(1,1), (2,2), (3,3), (4,4)$. Given pairs: $(1,2), (1,3)$. By symmetry: $(2,1), (3,1)$. Now we have $(1,2)$ and $(1,3)$, so by transitivity: $(2,3)$ (since $(2,1)$ and $(1,3)$ give $(2,3)$) and $(3,2)$ (since $(3,1)$ and $(1,2)$ give $(3,2)$). All pairs: $(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$. Total number of elements $= 10.$