Official Solution: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

First, we should determine the number of games played in this competition. We can count them in at least 2 different ways:

(1) Brute force. Name the 6 teams A, B, C, D, E, and F. 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.

(2) Combinatorics. We have a pool of 6 teams, and we want to count how many different pairs of teams (to play a game) we can select, without caring about order. Using either the anagram method or the formula for combinations, we get \(\frac{6!}{2!4!} = 15\) games.

Now, to find the maximum and minimum total points earned by all teams in the competition, we should notice that if one team wins and the other team loses, then 3 points total are earned (3 for the win and 0 for the loss). On the other hand, if the game ends in a draw, then only 2 points total are earned (1 by each team). So the maximum total points are earned if every game ends in a win/loss, and the minimum total points are earned if every game ends in a draw.

\(\text{Maximum} = 3 \times 15 = 45\) points.

\(\text{Minimum} = 2 \times 15 = 30\) points.

The difference between the maximum and the minimum is therefore \(45 - 30 = 15\).

Answer: A

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