Let's say that the people are:
A,B,C,D,E,F,G,H
The number of ways to select 2 people from 8 is 8C2, 2 from the remaining 6 is 6C2, 2 from the remaining 4 is 4C2 and 2 from the remaining 2 is (2C2).
So, the total number of ways should be (8C2).(6C2).(4C2).(2C2).
But, there is a problem in making teams like this. Lets look at a couple of examples of teams:
(A,B),(C,D), (D,E), (F,G)
(C,D),(F,G), (A,B), (D,E)
Notice that the above two set of teams are the same and should've been counted just once, but it will be counted unnecessarily again in the above method of selection. How many such repetitions are happening in each correct set of 4 teams? The 4 sets of teams will get arranged in 4! number of ways. Or, (8C2).(6C2).(4C2).(2C2) is each correct set of teams multiplied 4! times.
So, we will divide (8C2).(6C2).(4C2).(2C2) by 4! to get the correct answer, which is
( 28 X 15 X 6) / 24 = 105
