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Problem 268

Combinatorics Difficulty 2.3 Find the answer CEMC Pascal · Canada · 2016

A soccer team played three games. Each game ended in a win, loss, or tie. (If a game finishes with both teams having scored the same number of goals, the game ends in a tie.) In total, the team scored more goals than were scored against them. Which of the following combinations of outcomes is not possible for this team?

$\text{2 wins, 0 losses, 1 tie}$
$\text{1 win, 2 losses, 0 ties}$
$\text{0 wins, 1 loss, 2 ties}$
$\text{1 win, 1 loss, 1 tie}$
$\text{1 win, 0 losses, 2 ties}$

Your answer

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

Official solution

Since $x=3$ and $y=2x$ , then $y = 2\times 3 = 6$ .

Since $y=6$ and $z=3y$ , then $z = 3\times 6 = 18$ .

Therefore, the average of $x$ , $y$ and $z$ is $\dfrac{x+y+z}{3} = \dfrac{3+6+18}{3}=9$ .

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