We can mod out a symmetry by requiring $0 < a \leq b < c$. If $n'(p)$ denotes the number of such constrained $p$-smooth $abc$-triples, then $n(p) = 6n'(p) - 3$. This is due to a $6$-fold symmetry (by the anharmonic group
[7], which is isomorphic to $S_3$), which is only $3$-fold for the solution $1+1=2$.