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

We start with the set $$S=\{1,2,3\}$$ and require a relation that contains $$(1,2)$$ and $$(2,3)$$, is reflexive and transitive, but fails to be symmetric.

Reflexivity forces the relation to include $$(1,1)$$, $$(2,2)$$, and $$(3,3)$$.

Since $$(1,2)$$ and $$(2,3)$$ must be present, transitivity then requires $$(1,3)$$, so the mandatory core of the relation is $$\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}\;.$$

To violate symmetry, at least one of the reverse pairs $$(2,1)$$, $$(3,2)$$, $$(3,1)$$ must be missing while its forward counterpart is present.

Thus the optional pairs are $$(2,1),(3,1),(3,2)$$, each of which may be included or excluded provided transitivity is preserved and the relation remains non-symmetric. We now check all choices.

When none of the optional pairs is added, the relation is exactly the mandatory set. It is not symmetric because $$(1,2)$$ is in the relation but $$(2,1)$$ is not, and one checks that all transitivity conditions hold. Hence this choice is valid.

When only $$(2,1)$$ is added, one verifies that $$(2,1)$$ and $$(1,2)$$ yield $$(2,2)$$, that $$(2,1)$$ and $$(1,3)$$ yield $$(2,3)$$, and that $$(1,2)$$ and $$(2,1)$$ yield $$(1,1)$$, all of which are already in the relation. Since $$(2,3)$$ is present without $$(3,2)$$, the relation remains non-symmetric. Thus this choice is valid.

When only $$(3,1)$$ is added, transitivity with $$(3,1)$$ and $$(1,2)$$ would force $$(3,2)$$, which is not included, so this choice fails.

When only $$(3,2)$$ is added, one checks that $$(3,2)$$ and $$(2,3)$$ give $$(3,3)$$, that $$(3,2)$$ and $$(2,2)$$ give $$(3,2)$$, and that $$(1,3)$$ and $$(3,2)$$ give $$(1,2)$$, all of which already lie in the relation. Because $$(1,2)$$ appears without $$(2,1)$$, the relation remains non-symmetric, so this choice is valid.

Adding exactly $$(2,1)$$ and $$(3,1)$$ would force $$(3,2)$$ for transitivity (via $$(3,1)$$ and $$(1,2)$$), reducing to the case where all three optional pairs are present.

Adding exactly $$(2,1)$$ and $$(3,2)$$ forces $$(3,1)$$ (via $$(3,2)$$ and $$(2,1)$$), again yielding all three optional pairs.

When all three optional pairs $$(2,1),(3,1),(3,2)$$ are included, the relation becomes $$S\times S$$, which is symmetric and thus invalid.

Finally, adding exactly $$(3,1)$$ and $$(3,2)$$ satisfies transitivity without further additions and leaves $$(1,2)$$ unmatched by $$(2,1)$$, so the relation is non-symmetric and valid.

Altogether there are four valid choices (none of the optional pairs; only $$(2,1)$$; only $$(3,2)$$; or exactly $$(3,1)$$ and $$(3,2)$$), so the answer is \boxed{4}.