# Counting the number of terms coming from a Feynman Diagram

I am using the book QFT by Srednicki, Chapter 9. From the rules that he mentions, one can count the number of terms that correspond to the same Feynman diagram $$(t_i)$$. But for the case $$V=2, P=3$$, the diagrams are

For both of the diagrams I am getting the same number: $$(3!)^V$$ for the vertices and $$V!$$ for the permutations of the vertices and $$P!$$ for the permutations of all the propagators. This is $$t_1=t_2=(3!)^2\times2!\times3!=432$$. So, $$t_1+t_2=864$$. But $$t_1+t_2$$ should be equal to $$\frac{(2P)!}{(2P-3V)!}=720$$ in my understanding. What am I doing wrong here?

1. In OP's example we have $$V=2$$ cubic vertices $$Y$$ with in total 6 derivatives $$\frac{\delta}{\delta J}$$ attached, and $$P=3$$ propagators with in total 6 sources $$J$$ attached, cf. eq. (9.11) in Ref. 1. The derivatives and sources can be contracted in $$6!=720=288+432$$ ways, i.e. the number of permutations.

2. One may show that $$6\cdot 4\cdot 2\cdot 6=288$$ contractions$$^1$$ lead to the sunset diagram $$\theta$$ and $$6\cdot 4\cdot3\cdot6=432$$ contractions lead to the dumbbell diagram $$O\!\!-\!\!O$$.

3. If we divide with the normalization $$(3!)^V\cdot V! \cdot(2!)^P\cdot P!~=~(3!)^2\cdot 2! \cdot(2!)^3\cdot 3!~=~3456,$$ we get the reciprocal symmetry factors $$\frac{1}{12}$$ and $$\frac{1}{8}$$ for the 2 diagrams, respectively.

References:

1. M. Srednicki, QFT, 2007; equation (9.11) and figure 9.1. A prepublication draft PDF file is available here.

--

$$^1$$ Sketched proof for the sunset diagram $$\theta$$: Consider the 3 derivatives $$\frac{\delta}{\delta J}$$ on the 1st cubic vertex $$Y$$. The 1st derivative can be contracted in 6 ways. The 2nd derivative can only be contracted in 4 ways, since the 5th possibility would create a self-loop. Similarly, the 3rd derivative can only be contracted in 2 ways. Next consider the 3 derivatives $$\frac{\delta}{\delta J}$$ on the 2nd cubic vertex $$Y$$. Here all remaining contractions $$3!=6$$ work. In total there are $$6\cdot 4\cdot 2\cdot 6=288$$ contractions. $$\Box$$