Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 November Paper 52 Q7

A Level / CIE / S1

CIE 9709 2022 Nov Paper 52 Paper · Question 7

题目

Problem

(a) Find the number of different arrangements of the 9 letters in the word ALLIGATOR in which the two As are together and the two Ls are together.

(2)

(b) The 9 letters in the word ALLIGATOR are arranged in a random order.

Find the probability that the two Ls are together and there are exactly 6 letters between the two As.

(5)

(c) Find the number of different selections of 5 letters from the 9 letters in the word ALLIGATOR which contain at least one A and at most one L.

(3)

题目中文翻译

(a) 将单词 ALLIGATOR 中的 9 个字母排列,要求两个 A 相邻,且两个 L 相邻,求不同排列的个数。

(2)

(b) 将单词 ALLIGATOR 中的 9 个字母随机排列。

求两个 L 相邻且两个 A 之间恰好有 6 个字母的概率。

(5)

(c) 从单词 ALLIGATOR 的 9 个字母中选取 5 个字母,要求至少有一个 A 且至多有一个 L,求不同选取方式的种数。

(3)

原卷通用作答要求

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。

解答

(a)

解法一

思路

展开

AA和LL各为一个块,与另外五个字母形成七个不同对象。

答题过程

展开

The seven distinct objects AA, LL, I, G, T, O and R have

N=7!=5040N=7!=\boxed{5040}

arrangements.

(b)

解法一

思路

展开

两个A位置只能(1,8)或(2,9),外面唯一空位不能是L,否则LL无法相邻。选外面字母后,内部LL作块排列。

答题过程

展开

The As occupy positions (1,8)(1,8) or (2,9)(2,9), giving 2 possibilities.

In either case, the single position outside the As must contain one of I, G, T, O, R, giving 5 choices. The six letters between the As then consist of LL and four distinct letters. Treating LL as one block gives 5!5! arrangements.

The favourable number is therefore

2(5)(5!)=1200.2(5)(5!)=1200.

The total number of different arrangements is 9!2! 2!=90720\frac{9!}{2!\,2!}=90720. Hence

P=120090720=5378.\mathrm{P}=\frac{1200}{90720}=\boxed{\frac5{378}}.

解法二

思路

展开

另一条路线先在每组A位置下排列LL块及五个其他字母,再减去LL被一个A隔开的情形。

答题过程

展开

For each of the two A-position pairs, remove the As temporarily. The LL block and five other distinct letters have 6!6! arrangements.

When the As are reinserted, LL is split only if its two letters occupy the outside singleton position and the adjacent inside position. There are 5!5! such arrangements for each pair. Thus the favourable count is

2(6!−5!)=1440−240=1200.2(6!-5!)=1440-240=1200.

Therefore

P=2(6!−5!)9!/(2! 2!)=5378.\mathrm{P}=\frac{2(6!-5!)}{9!/(2!\,2!)}=\boxed{\frac5{378}}.

(c)

解法一

思路

展开

数不同字母组成时A不区分。先固定一A并去掉一L,剩下七种不同类型A、L、I、G、T、O、R中选4即可。

答题过程

展开

Set aside one A, which must be included. Remove one L, since at most one L is allowed.

The remaining seven distinct letter types are A, L, I, G, T, O, R. Choosing any four of them gives exactly one permitted selection of five letters, and every permitted selection is obtained once. Therefore

N=(74)=35.N=\binom74=\boxed{35}.

解法二

思路

展开

也可以按A有1或2个、L有0或1个分四类,其余从五个不同字母选。

答题过程

展开

The other five letters I, G, T, O, R are distinct. Classifying by the numbers of As and Ls gives

AsLsNumber of selections
1100(54)=5\binom54=5
2200(53)=10\binom53=10
1111(53)=10\binom53=10
2211(52)=10\binom52=10

Thus

N=5+10+10+10=35.N=5+10+10+10=\boxed{35}.