Yes. The thing to notice is that we can’t have three routes in the pattern (A,B,C), (A,B,D), (A,C,D), because in that case it’s impossible to fulfill the third condition: If the remaining ports are E and F, then we’ve left ourselves only one permissible route (not two) that visits A and E. That’s (A, E, F). Any other three-port route that includes A and E would “overload” another segment.

This means that if two of the routes are (A,B,C) and (A,B,D), then the second one that covers A,C must be (without loss of generality) (A,C,E). With that in mind the other routes can be worked out uniquely: (A,D,F), (A,E,F), (B,D,E), (B,E,F), (B,C,F), (C,D,E), (C,D,F).