(a) 将单词 ANDROMEDA 中的 9 个字母排列，要求任意两个辅音字母都不相邻，求不同排列的个数。（字母 D、M、N 和 R 是辅音字母，字母 A、E 和 O 不是辅音字母。）

(b) 将单词 ANDROMEDA 中的 9 个字母排列，要求两端各有一个 A，且两个 D 不相邻，求不同排列的个数。

从单词 ANDROMEDA 的 9 个字母中随机选取 4 个字母。

(c) 求所选字母中至少有一个 D 且恰好有一个 A 的概率。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

四个元音A、A、E、O先排列，产生五个空隙；五个辅音必须每个空隙恰放一个，辅音中两D相同。

Arrange the four vowels A, A, E, O in $\frac{4!}{2!}$ ways. This creates five gaps, including both end gaps.

The five consonants D, D, M, N, R must occupy these gaps one each, in $\frac{5!}{2!}$ ways, to prevent adjacent consonants. Hence

固定首尾A，从内部七字母的全部排列减去DD相邻的情况。

With an A at each end, the remaining letters D, D, N, R, O, M, E have $\frac{7!}{2!}$ arrangements.

If the Ds are together, their block and the other five distinct letters have $6!$ arrangements. Thus

另一条路线先排五个非D内部字母，再从六个空隙选两处放D。

Fix the As at the ends. Arrange N, R, O, M, E in $5!$ ways inside them, creating six gaps. Place the two identical Ds in two distinct gaps. Therefore

9个实际字母牌选择等概率。按恰一D或恰两D分类，同时恰一A。

There are $\binom94=126$ equally likely selections. Besides two As and two Ds, there are five other letters.

A permitted selection contains either one A, one D and two others, or one A, two Ds and one other. Thus

Hence

也可先选恰一A，另外三枚从七个非A字母选，再减去完全不含D的选择。

Choose the one A in $\binom21=2$ ways. There are $\binom73$ choices for the other three letters from the seven non-As, of which $\binom53$ contain no D. Therefore