still not 100% satisfied, so one last attempt:

A B C D

E F G H

I

Let's say those are the 9 guys, with A and B not to be on the same team.

Let's consider 9 cases, where each of 9 will be refs, then add all these 9 cases at the end. The reason i'm attempting this is because the 9th guy is the one that's making this problem confusing. if I can separate the problem into 9 cases, i can convert the problem into 9 separate 8C4 problems, which are super straight forward. 8 choose 4, and the left over 4 are the second team, so i don't even have to worry about the second choice. another way to think of it is 8C4 x 4c4.

Case 1: A is ref: 8C4/2! (or 8C4 x 4C4 / 2!) <-- same thing, depends on how you want to think of it.

Case 2: B is ref: 8C4/2! (or 8C4 x 4C4 / 2!) <-- same thing, depends on how you want to think of it.

we divide by 2! in the above 2 cases because if you choose 4 such that: B C D E are on the first team, the remaining 4 (F G H I) are on the other team. BUT choosing F G H I for the first team will give the exact same result - ie B C D E on one team, F G H I in the other. Since order doesn't matter, this is double counting. Exactly half of all possible choices will be duplicates, so we divide by 2.

Cases 3-9: C-I are refs (7 cases): A and B will be on separate teams. therefore: 6C3 (or 6C3 x 3C3 - depends on how you want to think of it)

7 such cases, so 7 x 6C3

we DON'T divide by 2! here because A and B are locked in, and since A and B are unique, choosing A C D E and A F G H (with i as ref, for example) are two separate cases.

answer: 8C4/2! + 8C4/2! + 7 x 6C3 = 210

i'm pretty sure 210 is the right answer. what answer were you getting herowise? i'm assuming you have the solutions, because how else do you know you're not getting it right etc. what's the answer?