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CIE 9709 2022 November Paper 51 Q6

A Level / CIE / S1

CIE 9709 2022 Nov Paper 51 Paper · Question 6

题目

Problem

A Social Club has 15 members, of whom 8 are men and 7 are women. The committee of the club consists of 5 of its members.

(a) Find the number of different ways in which the committee can be formed from the 15 members if it must include more men than women.

(4)

The 15 members are having their photograph taken. They stand in three rows, with 3 people in the front row, 5 people in the middle row and 7 people in the back row.

(b) In how many different ways can the 15 members of the club be divided into a group of 3, a group of 5 and a group of 7?

(3)

In one photograph Abel, Betty, Cally, Doug, Eve, Freya and Gino are the 7 members in the back row.

(c) In how many different ways can these 7 members be arranged so that Abel and Betty are next to each other and Freya and Gino are not next to each other?

(3)

题目中文翻译

一个社交俱乐部有 15 名成员,其中 8 名男性、7 名女性。俱乐部的委员会由其中 5 名成员组成。

(a) 若委员会中的男性人数必须多于女性人数,求从这 15 名成员中组成委员会的不同方式种数。

(4)

这 15 名成员要拍照。他们站成三排,前排 3 人、中排 5 人、后排 7 人。

(b) 将俱乐部的这 15 名成员分为 3 人、5 人和 7 人三组,有多少种不同的分组方式?

(3)

在其中一张照片中,后排的 7 名成员为 Abel、Betty、Cally、Doug、Eve、Freya 和 Gino。

(c) 若 Abel 与 Betty 相邻,且 Freya 与 Gino 不相邻,这 7 人有多少种不同的排列?

(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

5人中男性多于女性,只能是3男2女、4男1女、5男0女。

答题过程

展开 N= (83)(72)+(84)(71) +(85)(70)= 1176+490+56=1722.\begin{align*} N=&\,\binom83\binom72+\binom84\binom71\\[4mm] &\,\hspace{2pt}+\binom85\binom70\\[4mm] =&\,1176+490+56=\boxed{1722}. \end{align*}

解法二

思路

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另一条路线从全部5人委员会减去女性人数较多的三种构成。

答题过程

展开

The excluded compositions are 2 men and 3 women, 1 man and 4 women, and no men and 5 women. Hence

N= (155)−[(82)(73) +(81)(74)+(80)(75)]= 3003−(980+280+21)=1722.\begin{align*} N=&\,\binom{15}{5}-\biggl[\binom82\binom73\\[4mm] &\,\hspace{2pt}+\binom81\binom74+\binom80\binom75\biggr]\\[4mm] =&\,3003-(980+280+21)=\boxed{1722}. \end{align*}

(b)

解法一

思路

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三组人数不同,分别选前排3人与中排5人,剩下7人自动成组;不排组内位置也不除3!。

答题过程

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Choose the group of 3, then the group of 5 from the remaining twelve. The remaining seven form the last group. Thus

N=(153)(125)=455(792)=360360.N=\binom{15}{3}\binom{12}{5}=455(792)=\boxed{360360}.

(c)

解法一

思路

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Abel和Betty作为块,先数全部六对象排列,再减去Freya和Gino也作块的情况。

答题过程

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Treat Abel and Betty as a block with 2 internal orders. The block and the other five people have 6!6! arrangements.

If Freya and Gino are also together, the two blocks and the other three people have 5!5! arrangements, with 2 orders within each block. Hence

N=2(6!)−2(2)(5!)=1440−480=960.N=2(6!)-2(2)(5!)=1440-480=\boxed{960}.

解法二

思路

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也可以先排AB块及Cally、Doug、Eve,再将Freya与Gino放进两个不同空隙。

答题过程

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Arrange the Abel–Betty block, Cally, Doug and Eve in 4!4! ways, with 2 orders within the block. These four objects create five gaps.

Place Freya and Gino into two distinct gaps in 5×45\times4 ways. Thus

N=2(4!)(5)(4)=960.N=2(4!)(5)(4)=\boxed{960}.