Odds part 2

Not sure I understand the question....

   Take 10 rooms....put all team A members in those 10 rooms (2 persons to a room)

   Take the other 10 rooms and put all of the Team B members in those rooms.

We have 20 people of team A ,20 people of team B and 20 double rooms .How we can put the peaople so that in the rooms we only have people from the same team.

How many ways can 10 rooms be chosen from 20 rooms

20C10 = 184756

Team A can have these or they can have the other set. So that is 2 ways.

Now how many ways can I pair off 20 people?

Person 1 can go with anyone, 19 to chose from

Now there are 18 people left

Person 3 can go with anyone 17 to chose from

so there will be 16 left

and so on.

So that seem to be   19*17*15*13*11*9*7*5*3*1 = = 654729075

I really do not know if that is correct.

So I have

184756*2*654729075*654729075 158398768771746207705000 = 1.5839876877174621 * 10^23

Mmm that is a lot!

That is my best guess. (For the moment)

Thank you very much guys for help!