Symmetry factor of Feynman diagram in $\phi^4$ theory at second order 
I am trying to figure out the symmetry factor of the above diagram in the $\phi^4$ theory. The correct answer is $S=4$ (in that the Feynman diagram gets a multiplicative factor of $\frac{1}{4}$) but I get $S=8$ with the following reasoning.
For each closed loop, we can exchange the lines that start/end in the loop and belong to it. So, each loop gives a factor of $2$. Moreover, we can exchange the two vertices and simultaneously exchange the propagators of each loop. This gives an extra factor of $2$. I think it's the last bit that I'm getting wrong. Peskin and Schroeder write (p.93) that "A third possible type of symmetry is the equivalence of two vertices" but I don't think I understand precisely what this statement means. For example, why would the two vertices of this particular Feynman diagram not be equivalent? We can simply rename the integration variables and interchange them.
 A: Personally, I've always found the concept of symmetry factors more confusing than they're worth. I find it much easier and faster to simply count the number of contractions. The diagram you are looking at includes the fields
$$\phi_1 \phi_2 \phi_x\phi_x\phi_x\phi_x \phi_y\phi_y\phi_y\phi_y$$
$\phi_1$ can contract with 8 possible fields (any of the $\phi_x$ or $\phi_y$). Without loss of generality, we can call the field it contracts with $\phi_x$. $\phi_2$ must then contract with $\phi_y$ in 4 different ways. One of the remaining 3 $\phi_x$ (chosen in 3 ways) contracts with one of the remaining 3 $\phi_y$ (chosen in 3 different ways). All other contractions are then uniquely determined. The number of contractions is then
$$8\times 4\times 3\times 3$$
Putting in the factors of $\frac{1}{4!}$ for each vertex and a factor of $\frac{1}{2!}$ from the exponentiation, the overall factor in the Feynman diagram is
$$
\frac{1}{2!} \frac{1}{4! }\frac{1}{4!} 8\times 4\times 3\times 3 = \frac{1}{4} . 
$$
A: OP's main issue seems to be that the symmetry factor $S$ depends on whether the external legs are distinguishable ($S=4$) or indistinguishable ($S=8$), respectively.

*

*On one hand, P&S on p. 93 consider the distinguishable situation$^1$ where there are not integrated over the external positions/momenta.


*On the other hand, OP apparently considers the indistinguishable situation where there are attached sources $J$ to each external legs and integrated over the external positions/momenta.
P&S sentence "A third possible type of symmetry is the equivalence of two vertices" seems to refer to the $\frac{1}{n!}$ in Taylor expansion of the exponential function of an interaction term.
--
$^1$ Although P&S on p. 93 are not considering amputated diagrams it might be helpful to mention that the external legs are treated as distinguishable in an amputated diagram.
