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 $56 + 32 = \boxed{\text{(B) }88}$.

Solution 2

We can also figure out the number of games the 8 teams play within the conference with the formula n(n - 1)/2, which is used for these kind of match questions, where Team A playing Team B is the same as Team B playing Team A. In this formula, n is the number of teams. So we will do (8 X 7 / 2) X 2, as each team plays the other team twice. The 2s cancel out, to yield 8 X 7, which is 56 games. Then, we can just proceed as above to find the number of games the 8 teams play against 4 teams outside the conference, which is 8 X 4 = 32. Summing these values up yields 88, which is answer choice B.

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

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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