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 ______ -

We need to find the number of relations on $$A = \{1,2,3\}$$ that contain $$(1,2)$$ and $$(2,3)$$, are reflexive and transitive, but not symmetric.

Reflexivity requires the pairs $$(1,1), (2,2), (3,3)$$. Since $$(1,2)$$ and $$(2,3)$$ are given, transitivity forces us to include $$(1,3)$$. Thus the base set of pairs is $$\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$$.

Aside from these, the only remaining possible ordered pairs on $$A$$ are $$(2,1), (3,2), (3,1)$$. We examine which subsets of these can be added without violating transitivity and while ensuring the relation remains not symmetric.

If we add none of these extra pairs, the relation is already transitive and is not symmetric because $$(1,2)$$ is present but $$(2,1)$$ is absent. This case is valid.

If we add only $$(2,1)$$, then checking transitivity: $$(2,1)\circ(1,2)=(2,2)$$ ✓, $$(2,1)\circ(1,3)=(2,3)$$ ✓, and all other composites are already accounted for. Since $$(2,3)$$ is present without $$(3,2)$$, the relation remains not symmetric, so this case is valid.

If we add only $$(3,2)$$, then $$(3,2)\circ(2,3)=(3,3)$$ ✓, $$(1,3)\circ(3,2)=(1,2)$$ ✓, and all necessary closures hold. Because $$(1,2)$$ appears without $$(2,1)$$, the relation is not symmetric, so this case is valid.

If we add only $$(3,1)$$, transitivity would require $$(3,1)\circ(1,2)=(3,2)$$ to be present, but it is not. Therefore this case fails transitivity and is invalid.

If we add $$(2,1)$$ and $$(3,2)$$, then transitivity demands $$(3,2)\circ(2,1)=(3,1)$$, which is missing, so this case is invalid.

If we add $$(2,1)$$ and $$(3,1)$$, then $$(3,1)\circ(1,2)=(3,2)$$ must be present but is not, making this case invalid.

If we add $$(3,2)$$ and $$(3,1)$$, then $$(2,3)\circ(3,1)=(2,1)$$ is required and is missing, so this case is invalid.

If we add all three extra pairs $$(2,1),(3,2),(3,1)$$, the relation is transitive but becomes symmetric since every pair has its reverse, which is forbidden. Hence this case is invalid.

Out of the eight possibilities, only three yield relations that are reflexive, transitive, contain $$(1,2)$$ and $$(2,3)$$, and are not symmetric. Therefore, the answer is 3.