(a) 单词 SEYCHELLES 的 10 个字母共有多少种不同的排列？

(b) 将单词 SEYCHELLES 的 10 个字母排列成两个 S 之间恰有两个字母，且这两个字母之一为 C，共有多少种不同的排列？

(c) 将单词 SEYCHELLES 的 10 个字母排列成开头和末尾各有一个 S，且三个 E 不全部相连，共有多少种不同的排列？

从单词 SEYCHELLES 的 10 个字母中随机选出 5 个字母。

(d) 求所选的 5 个字母包含三个 E 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。

10个字母中S有2个、L有2个、E有3个。先排10个不同对象，再除去相同字母的内部排列。

There are two Ss, two Ls and three Es. Therefore the number of different arrangements is

两个S之间恰有两字母，所以位置相差3。先定两S的位置，再定C在中间两格中的哪一格；剩余7格放其余字母。

If the left S is in position $i$, the right S is in position $i+3$, so there are 7 choices for $i$.

C can occupy either of the two positions between them, giving 2 choices. The remaining seven positions contain Y, H, E, E, E, L, L, which have $\frac{7!}{3!\,2!}$ arrangements.

Thus the required number is

也可以按与C一起夹在两S之间的另一个字母分类，将这个四字母串作为块。另一个字母可以是Y、H、E或L；不同情况下剩余重复字母数不同。

Let $J$ be the other letter between the Ss. The block is SCJS or SJCS, with two internal orders.

Adding the four cases gives

固定首尾的S后，内部8个字母中L有2个、E有3个。先算全部内部排列，再减去EEE连成一块的排列。

Fix an S at each end. The eight internal letters have two identical Ls and three identical Es, giving $\frac{8!}{2!\,3!}$ arrangements.

If the three Es are together, treat EEE as one block. There are then six objects with two identical Ls, giving $\frac{6!}{2!}$ arrangements.

Thus the required number is

随机选择按10个实物字母计数。若包含全部三个E，只需再从其余7个字母中选两个，除以所有5字母选法。

Treat the physical letters as distinct when making the random selection.

There are $\binom{10}{5}=252$ equally likely selections. To include all three Es, select the other two letters from the seven non-E letters, giving $\binom72=21$ selections.

Hence

也可以按不放回的次序计算恰好抽到三个E。先指定一种三个E与两个非E的次序，再乘三个E可占的位置数。

Imagine drawing the five letters in order, without replacement.

For the order E, E, E, non-E, non-E, the probability is

There are $\binom53=10$ possible positions for the three Es, each with this probability. Thus