Problem 20 · 2006 AMC 8
Easy
Counting & Probability
round-robin
A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica (the sixth player) win?
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Answer: C — 2 games.
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Hint 1 of 2
Don't try to reconstruct who beat whom — you can't and you don't need to. Instead notice that every single game produces exactly ONE win, so total wins is a fixed, knowable number.
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Hint 2 of 2
Count the total number of games (each pair plays once), which equals total wins. Monica's wins are just that total minus everyone else's.
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Approach: every game makes exactly one win
- With 6 players each meeting every other once, the number of games is "choose 2 of 6" = (6 × 5)/2 = 15. Since there are no ties, that's exactly 15 wins handed out.
- The five named players account for 4 + 3 + 2 + 2 + 2 = 13 wins.
- Monica gets the rest: 15 − 13 = 2.
- You'll reuse this: in any round-robin with no ties, (total wins) = (total games) = "choose 2" of the players. The wins are conserved, so a single missing count is always "total minus the known." No game-by-game detective work needed.
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