AMC12 2016 B

AMC12 2016 B · Q20

AMC12 2016 B · Q20. It mainly tests Combinations, Casework.

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won 10 games and lost 10 games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A$?

一组队伍进行了循环赛，每支队伍与其他每支队伍都恰好比赛一次。每支队伍赢了 10 场、输了 10 场；没有平局。问有多少组三支队伍的集合 $\{A, B, C\}$ 满足：$A$ 战胜 $B$，$B$ 战胜 $C$，且 $C$ 战胜 $A$？

(A) 385 385

(B) 665 665

(D) 1140 1140

(E) 1330 1330

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): There must have been $10+10+1=21$ teams, and therefore there were $\binom{21}{3}=\frac{21\cdot20\cdot19}{6}=1330$ subsets $\{A,B,C\}$ of three teams. If such a subset does not satisfy the stated condition, then it consists of a team that beat both of the others. To count such subsets, note that there are $21$ choices for the winning team and $\binom{10}{2}=45$ choices for the other two teams in the subset. This gives $21\cdot45=945$ such subsets. The required answer is $1330-945=385$. To see that such a scenario is possible, arrange the teams in a circle, and let each team beat the $10$ teams that follow it in clockwise order around the circle.

答案（A）：一定有 $10+10+1=21$ 支队伍，因此三支队伍的子集 $\{A,B,C\}$ 的数量为 $\binom{21}{3}=\frac{21\cdot20\cdot19}{6}=1330$。如果某个子集不满足题设条件，那么它必然包含一支队伍击败了另外两支队伍。为统计这类子集，注意获胜队伍有 $21$ 种选择，而子集中另外两支队伍有 $\binom{10}{2}=45$ 种选择。因此这类子集共有 $21\cdot45=945$ 个。所求答案为 $1330-945=385$。为说明这种情形可实现，把队伍排成一个圆圈，并令每支队伍击败沿圆圈顺时针方向紧随其后的 $10$ 支队伍。

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