2014 AMC 8 Problems/Problem 16

Revision as of 14:20, 28 November 2014 by Nosaj (talk | contribs) (Solution)

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 $\text{(B)}$ is our answer.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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