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

We are given $A = \{1, 2, 3\}$ and the relation must contain $(1, 2)$ and $(1, 3)$.

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Since the relation is reflexive, $(a, a) \in R$ for every $a \in A$.

So $R$ must contain $(1,1),\ (2,2),\ (3,3)$.

Since the relation is symmetric, whenever $(a, b) \in R$, then $(b, a) \in R$.

$(1, 2) \in R \Rightarrow (2, 1) \in R$

$(1, 3) \in R \Rightarrow (3, 1) \in R$

So the minimum relation is:

$R_{\min} = \{(1,1),\ (2,2),\ (3,3),\ (1,2),\ (2,1),\ (1,3),\ (3,1)\}$

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The only remaining symmetric pair that can optionally be included is $(2, 3)$ and $(3, 2)$.

This gives two candidate relations:

$R_1 = R_{\min}$

$R_2 = R_{\min} \cup \{(2,3),\ (3,2)\}$

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Checking $R_1$ for transitivity:

$(2, 1) \in R_1$ and $(1, 3) \in R_1$, so we need $(2, 3) \in R_1$.

But $(2, 3) \notin R_1$, so transitivity fails.

$R_1$ is not transitive. ✓

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Checking $R_2$ for transitivity:

$R_2$ contains all 9 pairs in $A \times A$, making it the universal relation.

For any $(a, b)$ and $(b, c)$ in $R_2$, $(a, c)$ is always in $R_2$.

$R_2$ is transitive. ✗

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Only $R_1$ satisfies all conditions: reflexive, symmetric, contains $(1,2)$ and $(1,3)$, but not transitive.

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