 Solution

Hold the external endpoints fixed and distinguish the three Feynman diagrams by their left-to-right order in the source. Write $S$ for the divisor and $w=1/S$ for the multiplying combinatorial weight. Each line below includes external Feynman propagators; amputation leaves the same $w$.

Left: two-point bubble. There are two cubic interaction vertices and two parallel internal lines. Interchanging those lines preserves the labelled external endpoints, giving $S=2$. Direct Wick contraction counting gives two assignments of the external endpoints to the vertices, $3^2$ choices of their field slots, and $2!$ pairings of the remaining slots:With $\Delta ={\Delta }_F$ and external points $x_1,x_2$, its contribution is

Middle: three-point triangle. The three labelled external endpoints fix their three adjacent cubic interaction vertices, so no nonidentity graph automorphism remains. Counting gives $3!$ assignments of endpoints to vertices, $3^3$ choices of external field slots, and $2^3$ assignments of remaining slots to the two specified neighbours:Its contribution isThe apparent rotational symmetry of an unlabelled triangle does not survive fixing three distinct external endpoints.

Right: two-point graph with a chord. There are four cubic interaction vertices: the left and right attachment vertices $A,B$ and the two chord ends $C,D$. Its internal edges are $AC,AD,BC,BD,CD$. Reflection interchanges $C,D$, fixes $A,B$, and preserves the two external labels. This is the only nontrivial graph automorphism, so $S=2$.

For a separate Wick contraction count, choose the ordered external attachment vertices in $4\cdot 3$ ways and their field slots in $3^2$ ways. Assign their two remaining slots to $C,D$ in $2^2$ ways. At each of $C,D$, choose the slot joining $A$ and then the slot joining $B$ in $3\cdot 2$ ways; the last slots join each other. ThusIts contribution isConsequently the three multiplying weights are $\boxed{(1/2,1,1/2)}$, or divisors $\boxed{(2,1,2)}$. The central chord is a single internal line joining two vertices, not a vertex at the centre of the circle.