Let A = {1, 2, 3}. The number of relations on | vedclass

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(D) relation $R$ on $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Since $(1, 2) \in R$, symmetry requires $(2, 1) \in R$.For $R$ to be tra

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(D) relation $R$ on $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Since $(1, 2) \in R$, symmetry requires $(2, 1) \in R$.For $R$ to be transitive, since $(1, 2) \in R$ and $(2, 1) \in R$, we must have $(1, 1) \in R$ and $(2, 2) \in R$.Let $R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1)\}$. This relation is symmetric and transitive, but not reflexive on $A$ because $(3, 3) 
otin R_1$.If we include $(3, 3)$, the relation becomes $R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. This relation is symmetric, transitive, and reflexive.Any other relation containing $(1, 2)$ that is symmetric and transitive must contain $R_1$. If it does not contain $(3, 3)$, it is not reflexive. If it contains $(3, 3)$, it becomes reflexive.Thus, the only relation that is symmetric and transitive but not reflexive is $R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1)\}$.Therefore, there is only $1$ such relation.

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

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