Consider the one loop correction to a 2 -> 4 scattering process in phi4 theory.The only IPI/non snail contributions is that shown in the attachment. I have an automated package that will do all the feynman diagram generation for me and for this process it returns 15 diagrams, which means to say there are 15 inequivalent permutations of the external momenta. My question is basically how to derive this number?(adsbygoogle = window.adsbygoogle || []).push({});

If we draw the diagrams with 2 always incoming and 4 outgoing, then there are just 2! permutation of the incoming and 4! permutation of the outgoing. There are 6! permutation of all external lines naively and it seems that 6!/(2! x 4!) = 15, but I don't understand how this makes sense. Also, 5!/((2!)^3 = 15 which I can sort of make an argument for but I don't think it's a solid argument.

So,

a) what counts as an inequivalent diagram?

b) what is the correct combinatoric argument that gives rise to 15?

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# I Combinatorics of a phi4 interaction

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