How can the symmetry factor for a diagram be quickly determined and verified?

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Can someone check the symmetry factor I've found in the following diagram and verify it?
Do you know if there is any way to determine the SF of a diagram fast?

So I have \phi_x which can be contracted with 4 \phi_w : 4
Then \phi_y which can be contracted with 3 \phi_w : 3
Then I have 2 \phi_w which can be contracted with 4 \phi_u: 2x4
Then I have 3 \phi_u which can be contracted with 4 \phi_z: 3x4
The rest \phi_u get contracted together (only 2 left), whereas the \phi_z we get a factor of 2 since there are 2 possibilities to contract 3 fields.
Finally the last contraction gives just a factor 1 (no possible alternative choices).

So the result from contractions is 4x3x2 x4 x3 x2 x 4 = 4! x 4! x 4
The diagram is of order 3 (3 vertices) so there is a factor 1/3! and also the 1/4! from the coupling constant for \phi^4 theory.
So is it correct to say that the symmetry factor is afterall SF=16?
 

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Your factor of 2 (in blue) should be 3. From three fields, there are three possible distinct pairs to construct (AB, AC and BC).
But there is also a second point: there should be a factor of 1/4! for each vertex, so you should have 1/(4!)^3 instead of 1/4!
As for a shortcut: usually people drop the denominator and simply calculate the symmetry factor of the graph (the number of automorphisms in mathspeak). I think there is a nice discussion in Peskin and Schroeder if I recall correctly.
 
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In Peskin they say that they drop the 1/4! factor and write the vertex factor as \int d^4 z (-i \lambda) because they say that to a generic vertex has four lines coming in from four different places so the various contractions of \phi \phi \phi \phi is 4!...Also the n! from the Taylor expansion will cancel because of interchanging of vertices...
But I don't understand either of these explanations...
So in case I used (1/4!)3 I would have obtained the correct result?

A= \frac{16}{4! 4!}= \frac{1}{3! 3!} = \frac{1}{36}
So the symmetry factor is: 36?
 
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