Question

一场单打锦标赛有六名选手。每名选手与其他每名选手只比赛一次，没有平局。如果 Helen 赢了 $4$ 场，Ines 赢了 $3$ 场，Janet 赢了 $2$ 场，Kendra 赢了 $2$ 场，Lara 赢了 $2$ 场，那么 Monica 赢了几场？

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 win?

$0$

$1$

$2$

$3$

$4$

答案：C

知识点：组合

难度评级：1260

解答：

每场比赛恰好有一个胜者。六名选手共有 $6 \cdot 5 \div 2 = 15$ 场比赛，因此共有 $15$ 个胜场。

已知胜场数为 $4 + 3 + 2 + 2 + 2 = 13$ 所以 Monica 赢了 $15 - 13 = 2$ 场。

所以正确答案是 C。

In every match, there was exactly one winner. There are $6 \cdot 5 \div 2 = 15$ games and therefore $15$ wins.

There are already $4 + 3 + 2 + 2 + 2 = 13$ wins accounted for, so Monica won $15 - 13 = 2$ games.

Thus, C is the correct answer.

2006 AMC 8 第 20 题

答案：C

难度评级：1260

解答：

其他年份的第 20 题