Let $A = \{1, 2, 3\}$. Then, the number of relations containing (1, 2) and (1, 3), which are reflexive and symmetric but not transitive, is

1

The correct answer is Option (1) → 1

Set $A=\{1,2,3\}$, so all relations are subsets of $A\times A$.

A reflexive relation must contain $(1,1),(2,2),(3,3)$.

The relation must also contain $(1,2)$ and $(1,3)$, and by symmetry it must contain $(2,1)$ and $(3,1)$.

So far, the relation must include:

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

The remaining possible symmetric pairs are $(2,3)$ and $(3,2)$.

Let $x$ denote whether the pair $\{(2,3),(3,2)\}$ is included.

Case 1: $x=0$ (do not include $(2,3),(3,2)$). Then the relation is not transitive because $(2,1)$ and $(1,3)$ imply $(2,3)$ should be included but it is not. This gives 1 valid relation.

Case 2: $x=1$ (include both $(2,3),(3,2)$). Then the relation becomes fully connected and is transitive. This case is not allowed.

Hence only one relation satisfies all the conditions.

Final answer: $1$ relation