(a) 将单词 AMALGAMATE 中的 10 个字母排列，要求开头和末尾各有一个 M，且任意两个 A 都不相邻。求不同排列的个数。

(b) 将单词 AMALGAMATE 中的 10 个字母排列，要求两个 M 之间恰好有 3 个字母。求不同排列的个数。

从单词 AMALGAMATE 中的 10 个字母中选出 5 个字母。

(c) 求这 5 个字母包含至少一个 M 和至少两个 A 的不同选法数。

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。

固定首尾的M，再排G、L、T、E，产生五个内部空隙。四个A互不相邻，只能从五个空隙中选四个，每处放一个。

Fix an M at each end. Arrange G, L, T and E in $4!$ ways between them, creating five gaps, all between the Ms.

Choose four different gaps for the identical As, one per gap. Hence

两个M的位置相差4，左M有6种位置。剩下8格含四个相同的A。

If the left M is in position $i$, the right M is in position $i+4$, so $i$ can be 1 through 6.

The remaining eight letters contain four identical As, so the required number is

也可以将两个M与中间三个字母作为块。暂时标记四个A，选排块内字母后再排块和余下字母，最后消去标签。

Temporarily label the four As. Choose and order the three letters between the Ms in $\,{}_8\mathrm{P}_3$ ways.

The block M___M and the five remaining labelled letters have $6!$ arrangements. Removing the A labels divides by $4!$. Thus

不同选择不区分相同字母的实物身份。按M和A的数量分类，其余从四个不同字母中选。

The valid compositions are:

Therefore,

也可以先固定至少需要的一个M，按A有2、3、4个分类。其余可选字母为第二个M及G、L、T、E五种，每种最多一个。

Include one M first. The remaining letter types other than A are M, G, L, T and E.

With exactly two As, choose two of these five letters: $\binom52=10$ ways. With exactly three As, choose one: $\binom51=5$ ways. With four As, no further letter is needed: 1 way.

The Ms are identical, so fixing one introduces no choice of M labels. Thus