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

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

解答：

其他年份的第 22 题