2011 AMC 10A 第 22 题

Question

凸五边形

ABCDE

的每个顶点都要指定一种颜色。有

6

种颜色可选，并且每条对角线的两个端点必须颜色不同。有多少种不同的着色方法？

Each vertex of convex pentagon

ABCDE

is to be assigned a color. There are

6

colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

2520

2880

3120

3250

3750

答案：C

知识点：图论分类讨论排列

难度评级：1840

小提示：

按使用

5

、

4

或

3

种颜色分类。

Case on whether

5

,

4

, or

3

colors are used

大提示：

重复颜色只能出现在相邻顶点上。

Repeated colors can occur only on adjacent vertices

解答：

只有

3

种情况：所有顶点颜色都不同；恰有一对相邻顶点同色；或有

2

对相邻顶点同色，且两对使用不同颜色。

情况 $1$ ：所有顶点颜色都不同。

此时有

6! = 720

种着色方法。

情况 $2$ ：恰有一对相邻顶点同色。

选择各顶点颜色的方式有

\dfrac{6!}{2} = 360

种，而同色的相邻顶点对有

5

种选择。

因此这种情况共有

360 \cdot 5 = 1800

种着色方法。

情况 $3$ ：有两对相邻顶点分别同色。

不在任何同色对中的顶点有

5

种选择。再为两对和该单独顶点选择三种不同颜色，有

6 \cdot 5 \cdot 4 = 120

种，所以这种情况共有

120 \cdot 5 = 600

种着色方法。

所有情况合计

720 + 1800 + 600 = 3120

种。

所以正确答案是 C。

Note that there are only

3

cases: all the vertices are different, there is one pair of adjacent vertices with the same colors, or there are

2

pairs (each pair has a different color).

Case

1:

all vertices have different colors

This case just gives us

6! = 720

different colorings.

Case

2:

one pair of adjacent vertices has the same color

There are

\dfrac{6!}{2} = 360

ways to choose the colors for this case. There are then

5

options for the pair of vertices.

This gives us a total of

360 \cdot 5 = 1800

colorings for this case.

Case

3:

two pairs of adjacent vertices have the same color

There are

5

choices for the vertex that is not in a pair. There are then

6 \cdot 5 \cdot 4 = 120

choices for the colors. There are then a total of

120 \cdot 5 = 600

colorings for this case.

There are a total of

720 + 1800 + 600 = 3120

colorings for all the cases.

Thus, C is the correct answer.

2011 AMC 10A 第 22 题

答案：C

难度评级：1840

小提示：

大提示：

解答：

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