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CIE 9709 2023 June Paper 51 Q3

A Level / CIE / S1

CIE 9709 2023 June Paper 51 Paper · Question 3

题目

Problem

(a) Find the number of different arrangements of the 8 letters in the word COCOONED.

(1)

(b) Find the number of different arrangements of the 8 letters in the word COCOONED in which the first letter is O and the last letter is N.

(2)

(c) Find the probability that a randomly chosen arrangement of the 8 letters in the word COCOONED has all three Os together given that the two Cs are next to each other.

(3)

题目中文翻译

(a) 求单词 COCOONED 的 8 个字母的不同排列个数。

(1)

(b) 将单词 COCOONED 的 8 个字母排列,要求第一个字母为 O,最后一个字母为 N,求不同排列的个数。

(2)

(c) 随机选取单词 COCOONED 的 8 个字母的一种排列,已知两个 C 相邻,求三个 O 全部相邻的概率。

(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

两C与三O分别重复,其余字母不同。

答题过程

展开 N=8!2! 3!=3360.N=\frac{8!}{2!\,3!}=\boxed{3360}.

(b)

解法一

思路

展开

固定首O末N,内部剩两C两O与E、D。

答题过程

展开

After fixing O first and N last, the six remaining letters contain two Cs and two Os. Thus

N=6!2! 2!=180.N=\frac{6!}{2!\,2!}=\boxed{180}.

(c)

解法一

思路

展开

条件是CC相邻,分母仅数CC作块的排列;分子同时把CC与OOO都作块。

答题过程

展开

Given that the Cs are together, treat CC as one object. With O, O, O, N, E, D, there are 7!3!=840\frac{7!}{3!}=840 arrangements.

If the Os are also together, the five distinct objects CC, OOO, N, E, D have 5!=1205!=120 arrangements. Hence

P(OOO∣CC)=5!7!/3!=120840=17.\mathrm{P}(\text{OOO}\mid\text{CC})=\frac{5!}{7!/3!} =\frac{120}{840}=\boxed{\frac17}.