2021 AMC 12A Fall 第 22 题

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

小提示：

最终棋盘有三个

O

连成一行，且三个

X

不能自己也连成一行

The final board has three

O

s forming a line and three

X

s that must not themselves form a line

大提示：

按

O

的连线是行/列还是对角线分情况，因为这会改变

X

的放置中有多少会意外成行

Split into cases by whether

O

’s line is a row/column or a diagonal, since that changes how many

X

placements accidentally make a row

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

小提示：

大提示：

解答：

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