This question tests your knowledge of the Combination Formula, but it comes with a rare "twist" that most people don't realize. Here's a bit more information on that "twist":
With 10 players, the process of figuring out how many groups of 5 can be formed is pretty straight-forward....
10C5 = 10!/(5!5!) = 256 possible groups of 5
Once forming that first group of 5, the remaining 5 players would all be placed on the second team by default.
The 'twist' is that the two teams of 5 can "show up" in either order:
For example, if we call the two teams of 5 players: A,B,C,D,E and F,G,H,I,J
ABCDE vs. FGHIJ
is the SAME match-up as....
FGHIJ vs. ABCDE
So we are NOT allowed to count that matchup twice. This means we have to divide the 256 by 2.
GMAT assassins aren't born, they're made,
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