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Problem 566

AMC 10/12, early questions

Combinatorics Difficulty 3.6 Find the answer CEMC Fermat · Canada · 2020

Four teams play in a tournament in which each team plays exactly one game against each of the other three teams. At the end of each game, either the two teams tie or one team wins and the other team loses. A team is awarded 3 points for a win, 0 points for a loss, and 1 point for a tie. If $S$ is the sum of the points of the four teams after the tournament is complete, which of the following values can $S$ not equal?

$13$
$17$
$11$
$16$
$15$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Suppose that the four teams in the league are called W, X, Y, and Z.
Then there is a total of 6 games played:

W against X, W against Y, W against Z, X against Y, X against Z, Y against Z

In each game that is played, either one team is awarded 3 points for a win and the other is awarded 0 points for a loss (for a total of 3 points between the two teams), or each team is awarded 1 point for a tie (for a total of 2 points between the two teams).
Since 6 games are played, then the theoretical maximum number of points that could be awarded is $6 \cdot 3 = 18$ and the theoretical minimum number of points that can be awarded is $6 \cdot 2 = 12$ .
In particular, this means that it is not possible for the total number of points to be 11.
We can show that each of the possibilities from 12 to 18 points, inclusive, is actually possible.
Therefore, $S$ cannot equal 11.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.