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CIE 9709 2022 June Paper 51 Q2

A Level / CIE / S1

CIE 9709 2022 June Paper 51 Paper · Question 2

题目

Problem

There are 6 men and 8 women in a Book Club. The committee of the club consists of five of its members. Mr Lan and Mrs Lan are members of the club.

(a) In how many different ways can the committee be selected if exactly one of Mr Lan and Mrs Lan must be on the committee?

(2)

(b) In how many different ways can the committee be selected if Mrs Lan must be on the committee and there must be more women than men on the committee?

(4)

题目中文翻译

一个读书俱乐部有 6 名男性和 8 名女性。该俱乐部的委员会由其中 5 名成员组成。Lan 先生和 Lan 太太都是俱乐部成员。

(a) 如果 Lan 先生和 Lan 太太中必须恰有一人入选委员会,有多少种不同的选法?

(2)

(b) 如果 Lan 太太必须入选委员会,且委员会中的女性人数必须多于男性人数,有多少种不同的选法?

(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

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先选夫妻中哪一人入选,再从剩余12人中选4人。

答题过程

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Choose which one of Mr and Mrs Lan is included in 2 ways, then choose four of the other twelve members. Thus

N=2(124)=990.N=2\binom{12}{4}=\boxed{990}.

解法二

思路

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也可从全部五人委员会减去夫妻均入选和夫妻均不入选两类。

答题过程

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There are (145)\binom{14}{5} unrestricted committees. Committees containing both Mr and Mrs Lan number (123)\binom{12}{3}, and those containing neither number (125)\binom{12}{5}. Hence

N= (145)−(123)−(125)= 2002−220−792=990.\begin{align*} N=&\,\binom{14}{5}-\binom{12}{3}-\binom{12}{5}\\[4mm] =&\,2002-220-792=\boxed{990}. \end{align*}

(b)

解法一

思路

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固定Lan太太后剩7女6男,另选四人时需要至少两女,分2、3、4女三类。

答题过程

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Include Mrs Lan. Among the remaining seven women and six men, choose the other four members as follows:

Other womenMenNumber of committees
2222(72)(62)=315\binom72\binom62=315
3311(73)(61)=210\binom73\binom61=210
4400(74)=35\binom74=35

Thus

N=315+210+35=560.N=315+210+35=\boxed{560}.