Symmetry factor of certain 1-loop diagrams in $\phi^4$-theory

Question

I have to derive a formula for the symmetry factor of the diagrams of the form

in $\phi^4$-theory, where $\phi$ is a real scalar field. By symmetry factor I mean only the number of possible contractions, which lead to the same diagram (without the factor $1/n!$ for $n$th order of pertubation theory and without the factor $1/4!$ for each vertex from the Lagrangian).

So let $n$ be the number of external legs. For each diagram, we have a factor $(n/2)!$ from the interchangeability of the internal points. Furthermore, we get $(4!/2)^{n/2}$ to connect each pair of external lines to one of the vertices. What is left is the number of ways to connect the left internal lines, in order to get the circle...

In the first diagram, this gives a factor of $1$. In the 2nd diagram, we have a factor of $2$ and for the 3rd diagram, we have a factor of $2\cdot 2\cdot 2=4\cdot 2$. In a diagram with 4 pairs of external legs, we can simply see that we would get a factor of $6\cdot 4\cdot 2$. Therefore, we get a factor of $(n-2)!!$ for each diagram, for completing the circle.

In total, I find

$$S=(n/2)!\bigg (\frac{4!}{2}\bigg )^{n/2}(n-2)!!$$

However, I should have found

$$S=\bigg (\frac{4!}{2}\bigg )^{n/2}(n-1)!$$

according to the solution, which is clearly different from my expression. So, where is my error?

Answer 1

OP's exercise seems to conflate the number $n$ of external legs and the number $m=\frac{n}{2}$ of 4-vertices. The symmetry of an $m$-gon is $S(m\text{-gon})=2m=n$. The symmetry factor of the relevant 1-loop Feynman diagram is then $S=2^m S(m\text{-gon})=2^mn.$ For each vertex, there is $\begin{pmatrix}4\cr 2 \end{pmatrix}=6$ ways to choose a pair of halflines that participates in the loop. Since the order matters, this makes $2\times 6=12$ ordered pairs. We can order the vertices along the loop in $\frac{m!}{S(m\text{-gon})}=\frac{m!}{n}$ ways. Hence the sought-for number of contractions are $\#=12^m \frac{m!}{n}$. One may check that the resulting factor in the Feynman diagram becomes $\frac{\#}{m!(4!)^m}=\frac{1}{S}$ as it should.

Answer 2

Furthermore, we get $(4!/2)^{n/2}$ to connect each pair of external lines to one of the vertices

But if you consider the interchangeability of internal points then shouldn't it be just a factor of 2 for the way of choosing the connection between external line and vertex?