There are 2 triplets and 4 "single" friends.

We cannot choose both triplets for the same team since there can 
only by 5 on a team. 
 
We only need to choose one team, for the ones that are left
will automatically make up the other team.

We choose one of the triplets.  Then there are the combinations
of 4 single friends taken 2 at a time to go with that triplet.

The answer is C(4,2) = 6 ways

[Don't make the mistake of thinking there are 2 choices for the triplets,
since we only need to pick one triplet and assign two of the 4 "single"
players to be with that triplet, since the other team with the other 
triplet will automatically be chosen once we choose the 2 to go with that 
one triplet.]

Suppose one triplet is {A,B,C}, the other triplet (D,E,F}, and the
four "single" people are {G,H,I,J}

All 6 ways to divide the teams are:

1.  {A,B,C,G,H}  and the other team {D,E,F,I,J}
2.  {A,B,C,G,I}  and the other team {D,E,F,H,J}
3.  {A,B,C,G,J}  and the other team {D,E,F,H,I}
4.  {A,B,C,H,I}  and the other team {D,E,F,G,J}
5.  {A,B,C,H,J}  and the other team {D,E,F,G,I}
6.  {A,B,C,I,J}  and the other team {D,E,F,G,H}

Edwin