AMC10 2016 B

AMC10 2016 B · Q22

AMC10 2016 B · Q22. 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 $A$?

若干支队伍进行了一次循环赛（每支队伍与其他每支队伍恰好比赛一次）。每支队伍赢了 10 场、输了 10 场，且没有平局。问：有多少个由三支队伍组成的集合 $\{A,B,C\}$ 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，因此由三支队伍组成的子集 $\{A,B,C\}$ 的数量为 $\binom{21}{3}=\frac{21\cdot20\cdot19}{6}=1330$。如果这样的子集不满足题设条件，那么它一定包含一支击败另外两支的队伍。为了计数这类子集，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。因此这样的子集共有 $21\cdot45=945$ 个。所求答案是 $1330-945=385$。为说明这种情形可实现，可将队伍排成一圈，并让每支队伍按顺时针方向击败其后面的 10 支队伍。

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