Question

若干支队伍进行循环赛，每队与其他每队恰好比赛一次。每队赢 $10$ 场、输 $10$ 场，没有平局。有多少组三支队伍 $\{A,B,C\}$ ，满足 $A$ 击败 $B$ ， $B$ 击败 $C$ ，且 $C$ 击败 $A$ ？

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 $A?$

$385$

$665$

$945$

$1140$

$1330$

答案：A

知识点：图论补集计数组合

难度评级：2110

小提示：

每队与 $20$ 个对手比赛，所以共有 $21$ 支队伍，三队组总数为 $\binom{21}{3}=1330$ 。

Each team plays $20$ others, so there are $21$ teams and $\binom{21}{3}=1330$ triples in all

大提示：

非循环的三队组恰有一支队伍击败另外两支；数出这些并从总数中减去。

A non-cyclic triple has exactly one team that beats the other two; count those and subtract from the total

解答：

因为每队赢 $10$ 场、输 $10$ 场，所以共有 $21$ 支队伍，三队组总数为 $\binom{21}{3}=1330$ 。一个三队组不是循环关系，当且仅当其中某一队击败另外两队。选定这支胜队有 $21$ 种方式，再从被它击败的 $10$ 支队伍中选出 $2$ 支，所以非循环三队组共有 $21\cdot\binom{10}{2}=21\cdot45=945$ 个。因此循环三队组共有 $1330-945=385$ 个。

所以正确答案是 A。

Since each team won $10$ and lost $10,$ there are $21$ teams and $\binom{21}{3}=1330$ triples. A triple is not cyclic exactly when one team beats both others. Choosing that team ( $21$ ways) and $2$ of the $10$ teams it beat gives $21\cdot\binom{10}{2}=21\cdot45=945$ non-cyclic triples. Thus the cyclic triples number $1330-945=385.$

Thus, the correct answer is A.

2016 AMC 12B 第 20 题

答案：A

难度评级：2110

小提示：

大提示：

解答：

其他年份的第 20 题