Question

Central 高中与 Northern 高中进行十五子棋比赛。每所学校有三名选手，规则要求每名选手都与对方学校的每名选手比赛两局。比赛分六轮进行，每轮同时进行三局。共有多少种不同的赛程安排？

Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?

$540$

$600$

$720$

$810$

$900$

答案：E

知识点：有限制的排列多重集排列分类讨论

难度评级：2390

解答：

记 Central 的选手为 $A,B,C$ ，Northern 的选手为 $X,Y,Z$ 。

选手 $A$ 的六轮对手字符串中 $X,Y,Z$ 各出现两次，所以有 $\frac{6!}{2!2!2!}=90$ 种。

固定 $A$ 的字符串，例如 $XXYYZZ$ 。 $B$ 的字符串也必须含两个 $X,Y,Z$ ，且每一轮不能与 $A$ 的对手相同。

若 $B$ 的前两位为 $Y,Z$ 的某种顺序，中间两位必须为 $X,Z$ 的某种顺序，后两位必须为 $X,Y$ 的某种顺序，共 $2^3=8$ 种。另有 $YYZZXX$ 和 $ZZXXYY$ 两种，共 $10$ 种 $B$ 的字符串。

一旦 $A$ 和 $B$ 的赛程确定， $C$ 的赛程被迫确定。因此总数为 $90\cdot10=900$ 。

所以正确答案是 E。

Label Central's players $A,B,C$ and Northern's players $X,Y,Z$ .

Player $A$ 's six-round opponent string contains two each of $X,Y,Z$ , so it can be chosen in $\frac{6!}{2!2!2!}=90$ ways.

For a fixed $A$ -string, say $XXYYZZ$ , player $B$ 's string must also contain two each of $X,Y,Z$ and cannot match $A$ 's opponent in any position.

If $B$ 's first two entries are $Y,Z$ in either order, then the middle two entries must be $X,Z$ in either order and the last two must be $X,Y$ in either order, giving $2^3=8$ strings. The two remaining possibilities are $YYZZXX$ and $ZZXXYY$ , for $10$ total $B$ -strings.

Once $A$ and $B$ are scheduled, $C$ 's schedule is forced. Hence there are $90\cdot10=900$ schedules.

Thus, E is the correct answer.

2013 AMC 10A 第 24 题

答案：E

难度评级：2390

解答：

其他年份的第 24 题