David Griffiths: "Introduction to Elementary Particles", Second Revised Edition, formula (6.47) reads $ \frac { d \sigma } { d \Omega } = \left( \frac { \hbar c } { 8 \pi } \right) ^ 2 \frac { S | {\cal M} | ^ 2 } { ( E _ 1 + E _ 2 ) ^ 2 } \frac { | \mathbf { p } _ f | } { | \mathbf { p } _ i | } $ and expresses the differential cross section of a process $ 1 + 2 \rightarrow 3 + 4 $.

According to Griffiths, $ S $ is a statistical factor that corrects for double-counting when there are identical particles in the final state. In particlar, in Formula (6.47), $ S $ equals $ 1 $ if particles 3 and 4 are different and equals $ 1 / 2 $ if the particles are identical. Griffiths says no more about $ S $.

I have tried to Google for more information about $ S $ and exactly how to use it, but all I have found is that some web pages like https://www.nikhef.nl/~i93/Master/PP1/2017/Lectures/Lecture2017.pdf omit $ S $ and others like http://www.physics.utah.edu/~belz/phys5110/lecture23.pdf include it.

My question concerns the following thought experiment: Suppose $ A $, $ B $ and $ C $ are types of elementary particles and that $ A + B \rightarrow C + C $ is a possible process. Suppose we let a beam of $ A $-particles intersect with a beam of $ B $-particles. Suppose the luminosity of the collision (the thing to multiply $ \sigma $ by to get an event rate) is $ L $. Suppose we have a detector which detects $ C $-particles which are created by the process and travel in a particular direction within a small, spherical angle $ d \Omega $.

Now the question is: Is the expected event rate $ \frac { d \sigma } { d \Omega } L d\Omega $ or $ 2 \frac { d \sigma } { d \Omega } L d\Omega $. The argument for the latter would be that the detector could detect either of the $ C $'s in the final state. If the latter is correct, then we first have to multiply by $ S = 1 / 2 $ to account for identical particles and then multiply by $ 2 $, again to account for identical particles, and the two factors would cancel out in this particular case. Is that correct?