Round robin format

This is a Manhattan gmat question. It is posed on beatthegmat as well, with solution. What if i solve it in this particular way:

Let there be six teams as A,B,C,D,E and F.
Also, there will be total of six matches:

A plays each of the other teams once, so A plays 5 games. B also plays 5 games, but we’ve already counted 1 of those games (the game with A), so we have 4 “new” games. C also plays 5 games, but we’ve already counted 2 of those games (the games with A and with B), so we have 3 “new” games. Continuing, we get 5 + 4 + 3 + 2 + 1 = 15 games.

Now, Lets say, Team A plays with all the other teams, namely, B,C,D,E and F and wins.
So A gains 15 points.

Also, to maximise the difference, B loses the matches with C,D,E and F also along with A.
So, B has scores 0 point.

The difference between the maximum and the minimum score therefore is=15-0=15.

Where is my logic flawed?

parveen110 wrote:In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?

A. 15
B. 30
C. 45
D. 60
E. 75

Each game is played by a PAIR of teams.
Number of pairs that can be formed from 6 teams = 6C2 = (6*5)/(2*1) = 15.
Thus, there are a total of 15 games.

Every WIN yields a total of 3 points (3 points for the winner, 0 points for the loser).
Every DRAW yields a total of 2 points (1 point for each team playing).

The maximum number of points will be yielded if there are 15 wins:
15*3 = 45.
The minimum number of points will be yielded if there are 15 draws:
15*2 = 30.
Maximum - minimum = 45-30 = 15.

The correct answer is A.