I am reading this paper http://arxiv.org/abs/hep-th/0211102 and I would like to understand better about the branching rule $SO(6) \equiv SU(4) \rightarrow SU(3)$ used for eq. C.11 in the Appendix. I understand the branching rules of the 15 since it's the adjoint of $su(4)$ basically, but I don't really get which "20" decomposes as illustrated. So far (by some Young Tableaux computations) I found only inequivalent 20's with the following branchings:
$$ 20 \rightarrow 8+3+\bar 3+ 6 \\ 20' \rightarrow 10+6+3+1 \\ 20'' \rightarrow 6+6+8 $$
I am wondering whether I am missing some other inequivalent 20 or whether I am doing something wrong with the computation. So, how to understand the branching C.11 for that 20?