TITLE: How to count symmetry factors of Feynman diagrams? 
QUESTION [21 upvotes]: I have enough fears that this question might get struck down. Still let me try. 
I shall restrict myself to $\frac{\lambda \phi^4}{4!}$ perturbed real scalar quantum field theory and call as "symmetry factor" of a Feynman diagram to be the eventual number by which the power of $\lambda$ is divided in the final integral representation of the diagram. 
In that way the symmetry factor of the figure of eight vacuum bubble is 8 and of the "tadpole diagram" it is 2. 
One way to get this factor right is to count the number of ways the free arms in the "pre-diagram" can be contracted. But this is more like a cook-book rule than an understanding of how the factor comes. 
I believe the most conceptually correct way is to count for every diagram the number of  terms in the representation as a functional derivative of the path-integral which give that diagram. 
In that picture one has to argue that there were precisely $4!$ terms produced by the functional derivative which produced that figure of eight vacuum bubble. Which I can argue.
But for the tadpole diagram and the product of the vacuum bubble with the free-propagator, I can't find an argument. Like one has to be able to show that in the functional derivative picture there are $4!\times 3! \times (2!)^2$ terms corresponding to the tadpole diagram. 
Any help regarding how this counting is done or any general framework which helps compute these symmetry factors correctly?

REPLY [6 votes]: I think this question is mostly about mentally internalizing the interpretation of generating functions and the orbit-stabilizer theorem.  That is, if you can understand how the rules arise from a method that seems more natural to you, then the enumeration of automorphisms should appear less like arbitrary cookbook rules.
There is a good exposition of the symmetry factors in section 5 (page 25) of Alex Barnard's notes from Richard Borcherds's 2001 QFT class.  In order to distill the combinatorics of the diagrams, it gives a treatment of $\phi^4$ theory in zero spacetime dimensions, where the path integral is just an ordinary one dimensional integral: $$Z = \int_\mathbb{R} e^{-\phi^2/2 - \lambda \phi^4/4!}d\phi.$$
You can expand this integral as a series in polynomials times Gaussians, and observe that the $\frac{(2n)!}{n!2^n}$ in the identity $\int_\mathbb{R} \phi^{2n} e^{-\phi^2/2} d\phi = \frac{(2n)!}{n!2^n}\int_\mathbb{R} e^{-\phi^2/2}d\phi$ enumerates matchings of $2n$ labeled vertices.  If you ask for $k$ ($=\frac{n}{2}$) vertices to have valence 4, then there is a group of order $k!(4!)^k$ that permutes the vertices and their inputs, and this order divided by the order of the stabilizer of a particular isomorphism type is the order of the automorphism group.  By comparing coefficients, you find that a sum over graph isomorphism types, weighted by their automorphism groups, yields exactly the terms that come up in the path integral.
I imagine there may be a high-level explanation using species theory, but I'm not the right person to give it.