Question

Azar 和 Carl 玩井字棋。Azar 先在一个 $3$ 乘 $3$ 方格阵列中的某个格子放一个 $X$ ，然后 Carl 在剩余格子之一放一个 $O$ 。之后 Azar 在剩余格子之一放一个 $X$ ，依此类推，直到所有 $9$ 个格子都被填满，或者某一位玩家有 $3$ 个自己的符号横向、纵向或对角线连成一行；此时游戏立即停止，该玩家获胜。假设玩家随机落子，而不是试图遵循理性策略，并且 Carl 在放下第三个 $O$ 时赢得游戏。游戏结束后棋盘可能有多少种样子？

Azar and Carl play a game of tic-tac-toe. Azar places an $X$ in one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all $9$ boxes are filled or one of the players has $3$ of their symbols in a row — horizontal, vertical, or diagonal — whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O.$ How many ways can the board look after the game is over?

$36$

$112$

$120$

$148$

$160$

答案：D

知识点：分类讨论组合补集计数

难度评级：2270

解答：

Carl 在第三个 $O$ 时获胜，所以棋盘上有三个 $O$ 形成 $8$ 条线之一，另有三个 $X$ 在其余六个格子中。 $X$ 不能成一行（否则 Azar 已经先赢了）。

若 $O$ 的线是某一行或某一列（ $6$ 种选择），剩下六个格子包含两条完整线，所以有效 $X$ 放法数为 $\binom{6}{3} - 2 = 18$ 。若 $O$ 的线是对角线（ $2$ 种选择），剩下六个格子不包含完整线，有 $\binom{6}{3} = 20$ 种。

总数为 $6\cdot 18 + 2\cdot 20 = 108 + 40 = 148$ 。

所以正确答案是 D。

Carl wins on his third $O,$ so the board has three $O$ s forming one of the $8$ lines and three $X$ s in the other six cells. The $X$ s must not form a line (else Azar would have won first).

If the $O$ line is a row or column ( $6$ choices), the remaining six cells contain two full lines, so valid $X$ placements number $\binom{6}{3} - 2 = 18.$ If the $O$ line is a diagonal ( $2$ choices), the remaining six cells contain no full line, giving $\binom{6}{3} = 20.$

The total is $6\cdot 18 + 2\cdot 20 = 108 + 40 = 148.$

Thus, the correct answer is D.

2021 AMC 12A Fall 第 22 题

答案：D

难度评级：2270

解答：

其他年份的第 22 题