2014 AMC 8 Problems/Problem 16

Problem

The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?

$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

Solution

Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.

Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.

Therefore, the total number of games is $56+32 = \boxed{88}$ , so $\boxed{\text{(B)}}$ is our answer.

Video Solution 1 by OmegaLearn

https://youtu.be/Zhsb5lv6jCI?t=991

Video Solution (CREATIVE THINKING)

https://youtu.be/G2Akf39uwcE

~Education, the Study of Everything

Video Solution 3

https://youtu.be/Zhsb5lv6jCI?t=991

https://youtu.be/w7Y-iq_kEaY ~savannahsolver

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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