Question

一组队伍进行循环赛，每支队伍与其他每支队伍恰好比赛一次。每支队伍都赢了 $10$ 场、输了 $10$ 场，没有平局。有多少个三队集合 $\{A, B, C\}$ 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：1820

解答：

队伍总数为 $10+10+1=21$ 。因此三队集合总数为 $\binom{21}{3} = 1330$ 。

没有循环胜负的三队集合中，有一支队伍击败另外两支。选择这支队伍有 $21$ 种方法，再从它击败的十支队伍中选两支，有 $\binom{10}{2} = 45$ 种方法。因此非循环集合有 $21\cdot 45=945$ 个，循环集合数为 $1330-945=385$ 。

所以正确答案是 A。

The total number of teams is $10+10+1=21.$ The total number of sets is therefore $\binom{21}{3} = 1330.$

Now, we must subtract the total number of sets such that there is no cycle. This only happens if one team beats the other two teams. There are $21$ choices for the team that beat the other two and $\binom{10}{2} = 45$ ways to choose the teams they beat. Thus, the total of non-cycles is $21\cdot 45=945.$ This means the total number of cycles is $1330-945=385.$

Thus, the correct answer is A .

2016 AMC 10B 第 22 题

答案：A

难度评级：1820

解答：

其他年份的第 22 题