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

Here's a quick solution that doesn't require any "fancy" counting techniques.

Each team plays each of the other teams exactly once.
So, for example, Team A will play 7 games (i.e., it plays Teams B, C, D, E, F, G and H)
Team B will play 7 games (it plays Teams A, C, D, E, F, G and H)
Team C plays 7 games (it plays Teams A, B, D, E, F, G and H)
.
.
.
Team H plays 7 games.
So, we have 8 teams, and each team plays 7 games.
The total # of games = (7)(8) = 56

BUT, the answer here is NOT 56.
Notice that we've inadvertently counted every game twice.
For example, we counted A vs B as one game, and we counted B vs A as another game.
Since we've counted every game twice, we must take 56 and divide it by 2 to get 28

Answer: C

Cheers,
Brent

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

Another approach is to ask, "In how many different ways can we select two teams to play each other?"
Since the order of the selected teams does not matter (i.e., selecting A then B is the same as selecting B then A), we can use combinations.
We can select 2 teams from 8 teams in 8C2 ways = 28 ways.

Answer: C

If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent

Oops, just I realized that I solved question #133 from the Problem Solving section (not the Data Sufficiency section).

Maybe you could just post the actual question (in a new post??)

Cheers,
Brent