There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 64 possibilities (8*8) since each team has 8 players. So there are a total of 192 possibilities (64*3). I hope this method is easier for you to understand.
This justifies what answer for what question? Q & A:
There are 3 teams each with 8 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?
Options:
(A) 50
(B) 23
(C) 192
(D) 28
(E) 45
(C)