Let $A=\{1,2,3\}$. The number of relations on $A$, containing $(1,2)$ and $(2,3)$, which are reflexive and transitive but not symmetric, is
Solution:
Relation $R$ must contain reflexive pairs $\{(1,1), (2,2), (3,3)\}$ and given pairs $\{(1,2), (2,3)\}$.
For transitivity, $(1,2)$ and $(2,3) \in R \implies (1,3) \in R$.
$R_0 = \{ (1,1), (2,2), (3,3), (1,2), (2,3), (1,3) \}$
$\implies$ Reflexive & Transitive: Yes.
$\implies$ Symmetric: No (missing $(2,1)$).
$\implies$ Valid.
Now, Checking addition of remaining pairs $\{(2,1), (3,2), (3,1)\}$:
1. Add $(2,1)$ only $\implies$ $R_1 = R_0 \cup \{(2,1)\}$. Transitive. Not symmetric (missing $(3,2)$). $\implies$ Valid.
2. Add $(3,2)$ only $\implies$ $R_2 = R_0 \cup \{(3,2)\}$. Transitive. Not symmetric (missing $(2,1)$). $\implies$ Valid.
3. Add $(3,1)$ $\implies$ Requires adding $(2,1)$ and $(3,2)$ for transitivity, making it symmetric. $\implies$ Invalid.
$\implies$ Total valid relations = 3.
Ans. (3)