AMC12 2007 B

AMC12 2007 B · Q16

AMC12 2007 B · Q16. It mainly tests Counting with symmetry / Burnside (rare), Symmetry.

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

一个正四面体的每个面都被涂成红色、白色或蓝色。如果两个着色被认为不可区分，当且仅当两个带这些着色的全等四面体可以通过旋转使它们的外观完全相同。有多少种可区分的着色方式？

(A) 15 15

(B) 18 18

(D) 54 54

(E) 81 81

Answer

Correct choice: (A)

正确答案：（A）

Solution

A tetrahedron has 4 sides. We can ignore the rotation part of the question completely and focus on the colors. The ratio of the number of faces with each color must be one of the following: $4:0:0$, $3:1:0$, $2:2:0$, or $2:1:1$ The first ratio yields $3$ appearances, one of each color. The second ratio yields $3\cdot 2 = 6$ appearances, three choices for the first color, and two choices for the second. The third ratio yields $\binom{3}{2} = 3$ appearances since the two colors are interchangeable. The fourth ratio yields $3$ appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them. The total is $3 + 6 + 3 + 3 = 15$ appearances $\Rightarrow \mathrm{(A)}$

正四面体有4个面。我们可以完全忽略题目中关于旋转的部分，只关注颜色。每种颜色所占面的数量比必须是下列之一： $4:0:0$, $3:1:0$, $2:2:0$, 或 $2:1:1$ 第一种比例给出$3$种外观，每种颜色各一种。第二种比例给出$3\cdot 2 = 6$种外观：第一种颜色有三种选择，第二种颜色有两种选择。第三种比例给出$\binom{3}{2} = 3$种外观，因为两种颜色可以互换。第四种比例给出$3$种外观。第一种颜色有三种选择，而另外两种颜色可互换，因此只对应一种可区分的配对。总数为$3 + 6 + 3 + 3 = 15$种外观 $\Rightarrow \mathrm{(A)}$

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