2008 AMC 10B 第 22 题

Question

三颗红珠、两颗白珠和一颗蓝珠随机排成一行。相邻两颗珠子颜色都不同的概率是多少？

Three red beads, two white beads, and one blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?

\dfrac{1}{12}

\dfrac{1}{10}

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

答案：C

知识点：基本概率多重集排列有限制的排列

难度评级：1680

小提示：

可区分的颜色排列共有

\dfrac{6!}{3!\,2!}=60

种。

There are

\dfrac{6!}{3!\,2!}=60

distinguishable orderings

大提示：

先放三颗红珠使它们不相邻，再安放白珠和蓝珠。

First place the three red beads so no two are adjacent, then fit the white and blue beads

解答：

可区分的排列共有

\tfrac{6!}{3!\,2!}=60

种。三颗红珠必须占据不相邻的位置，可能的红珠位置为

\{1,3,5\},\{2,4,6\},\{1,3,6\}

和

\{1,4,6\}

。

对于

\{1,3,5\}

和

\{2,4,6\}

，剩余位置彼此不相邻，所以蓝珠可在

3

个位置中任意选择，共

3+3=6

种。对于

\{1,3,6\}

和

\{1,4,6\}

，剩余位置中有两个相邻，所以蓝珠必须隔开两颗白珠，共

2+2=4

种。

有效排列共

10

种，所以概率为

\tfrac{10}{60}=\tfrac16

。

所以正确答案是 C。

There are

\tfrac{6!}{3!\,2!}=60

distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are

\{1,3,5\},\{2,4,6\},\{1,3,6\},

and

\{1,4,6\}.

For

\{1,3,5\}

and

\{2,4,6\},

the remaining seats are mutually non-adjacent, so the blue bead can go in any of the

3,

giving

3+3=6.

For

\{1,3,6\}

and

\{1,4,6\},

two remaining seats are adjacent, so the blue must separate the whites, giving

2+2=4.

That is

10

valid orderings, so the probability is

\tfrac{10}{60}=\tfrac16.

Thus, the correct answer is C.

2008 AMC 10B 第 22 题

答案：C

难度评级：1680

小提示：

大提示：

解答：

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