AMC10 2021 A

AMC10 2021 A · Q25

AMC10 2021 A · Q25. It mainly tests Basic counting (rules of product/sum), Casework.

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?

有多少种方法将 3 个不可区分的红筹码、3 个不可区分的蓝筹码和 3 个不可区分的绿筹码放置在 $3 \times 3$ 网格的方格中，使得相同颜色的两个筹码不直接相邻，要么垂直要么水平？

(A) 12 12

(B) 18 18

(D) 30 30

(E) 36 36

Answer

Correct choice: (E)

正确答案：（E）

Solution

Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center. This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are $6$ ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{\textbf{(E)} ~36}$.

称不同颜色为 A、B、C。将这些颜色重新排列到这三个字母有 $3!=6$ 种方法，因此在网格中排列字母后必须乘以 6。假设 A 在中心。这意味着有 $4+2=6$ 种方法在网格中排列 A、B 和 C，并且有 6 种方法重新排列颜色。因此，总共有 $6\cdot6=36$ 种方法，即 $\boxed{\textbf{(E)} ~36}$。

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