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CIE 9709 2022 June Paper 51 Q1

A Level / CIE / S1

CIE 9709 2022 June Paper 51 Paper · Question 1

题目

Problem

(a) Find the number of different arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are together.

(2)

(b) Find the number of different arrangements of the 8 letters in the word DECEIVED in which the three Es are not all together.

(4)

题目中文翻译

(a) 求单词 DECEIVED 中 8 个字母的不同排列数,要求三个 E 全部相邻,且两个 D 相邻。

(2)

(b) 求单词 DECEIVED 中 8 个字母的不同排列数,要求三个 E 不全部相邻。

(4)

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

EEE与DD各作一个块,与C、I、V构成五个不同对象。

答题过程

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The five distinct objects EEE, DD, C, I and V have

N=5!=120N=5!=\boxed{120}

arrangements.

(b)

解法一

思路

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先数全部重复字母排列,再减去EEE全相邻的情况;不是要求所有E两两不相邻。

答题过程

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The total number of arrangements is 8!3! 2!=3360\frac{8!}{3!\,2!}=3360.

If all Es are together, the EEE block and D, D, C, I, V have 6!2!=360\frac{6!}{2!}=360 arrangements. Thus

N=8!3! 2!−6!2!=3360−360=3000.N=\frac{8!}{3!\,2!}-\frac{6!}{2!}=3360-360=\boxed{3000}.