(a) 求将单词 POSSESSION 的 10 个字母排列成四个 S 全部相连、而两个 O 不相邻的不同排列数。

(b) 求将单词 POSSESSION 的 10 个字母排列成 P 位于开头、N 位于末尾，且两个 O 之间恰有四个字母的不同排列数。

从单词 POSSESSION 的 10 个字母中选出 5 个字母。

(c) 求所选的 5 个字母包含至少两个 S 和至少一个 O 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。

把四个S作为一块，先算其全部排列，再减去两个O也相连的排列。两个O相同，计数时不能区分。

Treat SSSS as a single block. Together with P, O, O, E, I and N, there are seven objects, with two identical Os. Their arrangements number $\frac{7!}{2!}$.

If the Os are also together, treat OO as one block, leaving six distinct objects. Therefore the required number is

也可以先排S块和四个不同字母，再在六个空隙里选两个分别放O，保证两个O不相邻。

Arrange the five distinct objects SSSS, P, E, I and N in $5!$ ways.

They create six gaps, including the two end gaps. Choose two different gaps for the identical Os, one O in each. Thus the required number is

固定P和N后，中间8个位置放两个O、四个S、E和I。两O相距5个位置，左O只能在中间8格的前3格；其余6格有四个相同的S。

Fix P at the beginning and N at the end. Number the eight internal positions from 1 to 8.

If the left O occupies position $i$, the right O must occupy position $i+5$. Thus $i$ can be 1, 2 or 3.

The six remaining positions contain S, S, S, S, E and I, giving $\frac{6!}{4!}$ arrangements. Hence the required number is

另一条路线是将两O及中间四个字母作为一个块。先暂时区分四个S，再选择块内四个字母并排列块与剩余字母，最后去掉S的标签。

Temporarily label the four Ss. After fixing P and N, the letters available between them are two identical Os and six distinct labelled letters.

Choose and order four of the six labelled letters between the Os in $\,{}_6\mathrm{P}_4$ ways. Treat O____O as a block. Together with the two remaining labelled letters, this gives $3!$ arrangements.

Removing the S labels divides the count by $4!$. Thus

题目求不同选取而不是概率，相同字母不按实物区分。按S和O的个数分类，余下只能从P、E、I、N四个不同字母中选。

The available letters are four Ss, two Os and the four distinct letters P, E, I, N.

With at least two Ss and at least one O, the possible numbers of Ss and Os in a five-letter selection are:

The repeated Ss and Os do not create additional different selections. The total is