Problem Solving

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

(A) 15
(B) 16
(C) 28
(D) 56
(E) 64

Guys, please help me with this one. I have difficulty in undertanding the wording of this problem.

Thanks!

IMO C (28)

You have 8 teams (A,B,C,D,E,F,G,H) that play against one another.
A vs. B
A vs. C
A vs. D
A vs. E
A vs. F
A vs. G
A vs. H

Since no team can play against itself, A plays 7 games.
B plays also 7 games, but now we have counted the game A vs. B and B vs. A which is the same.
Therefore we must substract this game once.
You can do this with all combinations and you end up with: 7+6+5+4+3+2+1=28

It seems to take quite some time, but actually it doesn't, only about 30 seconds after you figured out what to do.

Again it's a combination problem, take total combination of 2 team out of 8 team, i.e 8C2=>8*7/2=>28

What if instead of 2 teams, 3 teams play a game then how do we find total number of games.

magicmover wrote:What if instead of 2 teams, 3 teams play a game then how do we find total number of games.

8!/(5!)(3!) = 56