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

The correct option is 1.

Let us find the relations on the set $A=\{1,2,3\}$ that satisfy the given conditions.
A relation $R$ on $A$ is:
1. Reflexive if $(a,a)\in R$ for all $a\in A$. Therefore, $R$ must contain:
$\left\{\right(1,1),(2,2),(3,3\left)\right\}$.
2. Symmetric if $(a,b)\in R\Rightarrow (b,a)\in R$.

The question specifies that the relation $R$ must contain $(1,2)$ and $(1,3)$.
Since $R$ is symmetric, it must also contain their symmetric counterparts:
$(2,1)$ (because $(1,2)\in R$) and $(3,1)$ (because $(1,3)\in R$).

Thus, the minimum relation $R$ containing $(1,2)$ and $(1,3)$ that is both reflexive and symmetric is:
$R=\left\{\right(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1\left)\right\}$.

Now, let us check the transitivity of this relation $R$:
We have $(2,1)\in R$ and $(1,3)\in R$.
If $R$ were transitive, it would also have to contain $(2,3)$. But $(2,3)\notin R$.
Thus, this relation $R$ is reflexive and symmetric, but not transitive.

If we try to add any other elements to $R$ to form a different relation, the only remaining elements from $A\times A$ that are not yet in $R$ are $(2,3)$ and $(3,2)$.
If we add $(2,3)$, we must also add $(3,2)$ to maintain symmetry.
Doing so yields the universal relation $A\times A$, which is transitive.

Therefore, there is exactly 1 relation that is reflexive and symmetric but not transitive containing the pairs $(1,2)$ and $(1,3)$.