从单词 RETRIEVES 的 9 个字母中随机选出 3 个字母。

(a) 求所选字母中至少有一个 E 的概率。

(b) 求将单词 RETRIEVES 的 9 个字母排列成两端各有一个 R，且三个 E 不全部相连的不同排列数。

(c) 求将单词 RETRIEVES 的 9 个字母排列成两个 R 之间恰有三个字母的不同排列数。

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。

从9个字母位置中等可能地选3个。先算一个E也没有的概率，再用1减去。

There are three Es and six other letters. Treating the physical letters as distinct when making the random selection, there are $\binom93=84$ equally likely selections.

The number with no E is $\binom63=20$. Hence

也可以逐次考虑不放回取字母。每次没有取到E的条件概率分别为九分之六、八分之五、七分之四，再取补集。

The probability that all three selected letters are not E is

Thus

另一条组合路线是按恰有1、2、3个E分类。注意这些分类互斥，分子相加后再除以总选法数。

The numbers of selections with exactly one, two or three Es are respectively

Therefore,

固定两端的R，内部剩下三个相同的E和四个不同字母。先算所有内部排列，再减去三个E连成一块的排列。

With an R fixed at each end, the seven internal letters are E, E, E, T, I, V, S. Their total number of arrangements is

If all three Es are together, treat EEE as one object. Along with T, I, V and S, there are five distinct objects, giving $5!=120$ arrangements.

The required number is therefore

也可以先排T、I、V、S，然后在五个空隙中放三个E。要避免全部相连，只需考虑分成“2个加1个”或三个分别放置。

Fix the Rs at the two ends and arrange T, I, V, S in $4!$ ways. These four letters create five gaps, all between the two Rs.

For the Es not to be all together, either:

• place two Es in one gap and one E in another: $5\times4=20$ ways;
• place one E in each of three gaps: $\binom53=10$ ways.

Hence the required number is

先选两个R的位置。左边的R只能在前5个位置之一，另一个R的位置随之确定；剩下7个位置放其余字母，三个E相同。

Number the positions from 1 to 9. If the left R is in position $i$, the right R must be in position $i+4$.

Thus $i$ can be $1,2,3,4$ or $5$, giving five choices for the pair of R positions. The remaining seven letters can be arranged in $\frac{7!}{3!}$ ways, so the required number is

也可以将两个R和中间的三个字母合成一块。暂时将三个E加上标签，使另外7个字母各不相同；排好以后再除以三个E的标签交换数。

Temporarily label the Es as $E_1,E_2,E_3$.

Choose and order the three letters between the Rs in $\,{}_7\mathrm{P}_3$ ways. Treat the resulting string R___R as one block. Together with the four remaining letters, there are five distinct objects, which can be arranged in $5!$ ways.

Removing the labels from the Es divides the count by $3!$. Hence the number of different arrangements is