Problem 16 · 2014 AMC 8
Medium
Counting & Probability
round-robinavoid-double-counting
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?
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Answer: B — 88 games.
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Hint 1 of 2
The trap is double-counting: a conference game is shared by two MSE teams, but a non-conference game touches only one. Count the two kinds with different rules.
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Hint 2 of 2
Conference games are pairs of teams (each pair twice); non-conference games belong to a single team, so just add 8 × 4 with no halving.
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Approach: count conference pairs × 2, plus non-conference (no double-count)
- Conference: number of team-pairs is C(8, 2) = 28, and each pair plays twice (home and away) ⇒ 28 × 2 = 56 games.
- Non-conference: each of the 8 teams plays 4, and the opponent is outside MSE, so every such game is counted once — 8 × 4 = 32 games.
- Total: 56 + 32 = 88.
- Why this transfers: when both participants are inside your group, you divide by 2 (or use C(n,2)) to avoid double-counting; when only one participant is inside, you don't. Always ask "is this game shared?"
Another way — count by home games (each conference game has exactly one host):
- Every conference game is hosted by exactly one team, and each team hosts the other 7 once: 8 × 7 = 56 conference games — no double-counting since "host" is unique.
- Add the 8 × 4 = 32 non-conference games: 56 + 32 = 88.
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