AMC10 2008 B

AMC10 2008 B · Q22

AMC10 2008 B · Q22. It mainly tests Combinations, Probability (basic).

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?

三个红珠子、两个白珠子和一个蓝珠子随机排成一行。邻近的珠子没有相同颜色的概率是多少？

(A) \frac{1}{12} \frac{1}{12}

(B) \frac{1}{10} \frac{1}{10}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): There are $6!/(3!2!1!)=60$ distinguishable orders of the beads on the line. To meet the required condition, the red beads must be placed in one of four configurations: positions 1, 3, and 5, positions 2, 4, and 6, positions 1, 3, and 6, or positions 1, 4, and 6. In the first two cases, the blue bead can be placed in any of the three remaining positions. In the last two cases, the blue bead can be placed in either of the two adjacent remaining positions. In each case, the placement of the white beads is then determined. Hence there are $2\cdot 3+2\cdot 2=10$ orders that meet the required condition, and the requested probability is $\frac{10}{60}=\frac{1}{6}$.

答案（C）：线上珠子的不同排列数为 $6!/(3!2!1!)=60$。为满足题设条件，红珠必须放在以下四种位置配置之一：第 1、3、5 位；第 2、4、6 位；第 1、3、6 位；或第 1、4、6 位。前两种情况下，蓝珠可以放在剩下的三个位置中的任意一个。后两种情况下，蓝珠只能放在剩下的两个相邻位置中的任意一个。每种情况下，白珠的位置随之确定。因此满足条件的排列共有 $2\cdot 3+2\cdot 2=10$ 种，所求概率为 $\frac{10}{60}=\frac{1}{6}$。

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