2016 AMC 10B 第 22 题

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

小提示：

先数所有三队集合，再减去没有循环胜负的集合。

Count all triples of teams, then subtract transitive triples

大提示：

每支队伍作为最强队，会出现在

\binom{10}{2}

个传递型三队集合中。

Each team is the top team in

\binom{10}{2}

transitive triples

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

小提示：

大提示：

解答：

其他年份的第 22 题