题目

Problem

A group of 15 friends visit an adventure park. The group consists of four families.

Mr and Mrs Kenny and their four children
Mr and Mrs Lizo and their three children
Mrs Martin and her child
Mr and Mrs Nantes

The group travel to the park in three cars, one containing 6 people, one containing 5 people and one containing 4 people. The cars are driven by Mr Lizo, Mrs Martin and Mr Nantes respectively.

(a) In how many different ways can the remaining 12 members of the group be divided between the three cars?

(3)

The group enter the park by walking through a gate one at a time.

(b) In how many different orders can the 15 friends go through the gate if Mr Lizo goes first and each family stays together?

(3)

In the park, the group enter a competition which requires a team of 4 adults and 3 children.

(2)

(d) In how many ways can the team be chosen so that at least one of Mr Kenny or Mr Lizo is included?

(3)

题目中文翻译

一群 15 位朋友前往一个冒险乐园。他们由以下四个家庭组成。

Kenny 先生、太太及他们的四个孩子
Lizo 先生、太太及他们的三个孩子
Martin 太太及她的一个孩子
Nantes 先生和太太

他们乘三辆车前往乐园，一辆坐 6 人，一辆坐 5 人，一辆坐 4 人，分别由 Lizo 先生、Martin 太太和 Nantes 先生驾驶。

(a) 将其余 12 人分配到这三辆车中，有多少种不同的方法？

(3)

他们一个接一个地走过一道门进入乐园。

(b) 如果 Lizo 先生第一个通过门，且每个家庭的成员保持相邻，这 15 位朋友有多少种不同的过门顺序？

(3)

在乐园里，他们参加一项比赛，需要组成一支有 4 名成人和 3 名儿童的队伍。

(2)

(d) 选出队伍，要求 Kenny 先生和 Lizo 先生中至少有一人入选，有多少种选法？

(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

三位司机已固定，剩余座位分别5、4、3，依次选人且不排座位。

答题过程

展开

After the three drivers are assigned, the cars have 5, 4 and 3 remaining places respectively. Choose the passengers for the first car, then the second; the others fill the third:

N=\binom{12}{5}\binom74\binom33=792(35)=\boxed{27720}.

(b)

解法一

思路

展开

Lizo家必须是首块且Lizo先生在块首；其余三家庭可任意排列，各家庭内部另排列。

答题过程

展开

Mr Lizo goes first, so his family is the first block. The other four members of his family can be ordered in $4!$ ways.

The remaining three family blocks can be ordered in $3!$ ways, and their internal orders number $6!$ , $2!$ and $2!$ . Thus

N=4!\,3!\,6!\,2!\,2!=\boxed{414720}.

(c)

解法一

思路

展开

儿童只有三家，必须Kenny、Lizo、Martin各选一人；成人从七人选四。

答题过程

展开

There are 7 adults and 8 children. To have children from three different families, choose one of the four Kenny children, one of the three Lizo children and the Martin child. Hence

N=\binom74(4)(3)(1)=\boxed{420}.

(d)

解法一

思路

展开

本小题不继承(c)的儿童不同家庭限制。按仅Kenny、仅Lizo、两人皆有分类。

答题过程

展开

The three children may now be any three of the eight children. Classify the four adults by the inclusion of Mr Kenny and Mr Lizo:

Included	Adult selections
Mr Kenny but not Mr Lizo	$\binom53=10$
Mr Lizo but not Mr Kenny	$\binom53=10$
Both	$\binom52=10$

Thus

\begin{align*} N=&\,\biggl[\binom53+\binom53+\binom52\biggr]\binom83\\[4mm] =&\,30(56)=\boxed{1680}. \end{align*}

解法二

思路

展开

也可从所有队伍减去两位先生均未入选的队伍。

答题过程

展开

The unrestricted number of teams is $\binom74\binom83$ . If neither Mr Kenny nor Mr Lizo is included, all four adults must come from the other five. Therefore

\begin{align*} N=&\,\biggl[\binom74-\binom54\biggr]\binom83\\[4mm] =&\,(35-5)(56)=\boxed{1680}. \end{align*}

解法三

思路

展开

第三种路线用容斥：含Kenny与含Lizo相加，再减重复计算的两人皆有。

答题过程

展开

There are $\binom63\binom83$ teams containing Mr Kenny, and the same number containing Mr Lizo. Teams containing both number $\binom52\binom83$ and have been counted twice. Hence

\begin{align*} N=&\,\biggl[2\binom63-\binom52\biggr]\binom83\\[4mm] =&\,1120+1120-560=\boxed{1680}. \end{align*}

CIE 9709 2022 June Paper 53 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程

解法二

思路

答题过程

解法三

思路

答题过程