Question

有多少种方法把 $3$ 枚不可区分的红色筹码、 $3$ 枚不可区分的蓝色筹码和 $3$ 枚不可区分的绿色筹码放入一个 $3 \times 3$ 方格的格子中，使得任意两个同色筹码都不在上下或左右方向相邻？

How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?

$12$

$18$

$24$

$30$

$36$

答案：E

知识点：有限制的排列分类讨论

难度评级：1820

小提示：

先选定中心格的颜色，以及被这种颜色占据的两个角格。

First choose the color in the center and the two corners occupied by that color

大提示：

这三枚筹码放好后，另外两种颜色的位置就被确定，至多只差一次互换。

Once those three chips are placed, the other two colors are forced up to interchange

解答：

中心格的颜色有 $3$ 种选法。这种颜色的另外两枚筹码都不能放在边中格，因为这些格子与中心格相邻。因此它们只能占据四个角格中的两个，共有 $\binom42=6$ 种选法。

无论这两个角格是哪一种情形——处于对角，还是处于同一边的两端——相邻条件都会把剩下两种颜色的位置唯一确定，至多相差二者的互换。因此对中心颜色及其两个角格的每一种选择，都恰有 $2$ 种补全方式。总数为 $3\binom42\cdot2=36\text{。}$

所以正确答案是 E。

Choose the center color in $3$ ways. Its other two chips cannot occupy any edge-middle square, because those squares are adjacent to the center. Thus they must occupy two of the four corners, which can be chosen in $\binom42=6$ ways.

For either possible corner pattern—two opposite corners or two corners on the same side—the adjacency conditions force the remaining two colors up to interchanging them. Hence there are $2$ completions for each choice of the center color and its two corners. The total is $3\binom42\cdot2=36.$

Thus, E is the correct answer.

2021 AMC 10A Spring 第 25 题

答案：E

难度评级：1820

小提示：

大提示：

解答：

其他年份的第 25 题