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CIE 9709 2023 November Paper 51 Q6

A Level / CIE / S1

CIE 9709 2023 Nov Paper 51 Paper · Question 6

题目

Problem

In a restaurant, the tables are rectangular. Each table seats four people: two along each of the longer sides of the table (see diagram). Eight friends have booked two tables, XX and YY. Rajid, Sue and Tan are three of these friends.

(a) The eight friends will be divided into two groups of 4, one group for table XX and one group for table YY.

Find the number of ways in which this can be done if Rajid and Sue must sit at the same table as each other and Tan must sit at the other table.

(3)

When the friends arrive at the restaurant, Rajid and Sue now decide to sit at table XX on the same side as each other. Tan decides that he does not mind at which table he sits.

(b) Find the number of different seating arrangements for the 8 friends.

(3)

As they leave the restaurant, the 8 friends stand in a line for a photograph.

(c) Find the number of different arrangements if Rajid and Sue stand next to each other, but neither is at an end of the line.

(4)
题目中文翻译

一家餐厅的桌子都是长方形。每张桌子可坐四人，桌子的两条长边各坐两人（见图）。八位朋友预订了 XX 和 YY 两张桌子。Rajid、Sue 和 Tan 是其中三人。

(a) 将这八位朋友分成两个 4 人小组，一组坐 XX 桌，另一组坐 YY 桌。

如果 Rajid 和 Sue 必须坐在同一张桌子，而 Tan 必须坐在另一张桌子，共有多少种分组方法？

(3)

朋友们到达餐厅后，Rajid 和 Sue 改为决定坐在 XX 桌的同一侧。Tan 表示坐在哪张桌子都可以。

(b) 求这 8 位朋友不同的座位安排数。

(3)

离开餐厅时，这 8 位朋友站成一排拍照。

(c) 如果 Rajid 和 Sue 相邻站立，但两人都不站在队列两端，共有多少种不同的排列？

(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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Rajid与Sue同桌，先选另两位同桌者，不能选Tan；再选这组坐哪张桌。

答题过程

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Choose two of the five friends other than Rajid, Sue and Tan to join Rajid and Sue, in (52)\binom52 ways. The other four, including Tan, form the other group.

There are two choices of table for Rajid and Sue’s group. Hence

N=2(52)=20.N=2\binom52=\boxed{20}.

(b)

解法一

思路

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X桌同一长边有2种选择，Rajid与Sue有2种座位顺序，剩六人填剩六席。

答题过程

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Choose one of the two long sides of table X for Rajid and Sue, and arrange them there in 2 ways. The other six friends occupy the six remaining distinct seats in 6!6! ways. Thus

N=2(2)(6!)=2880.N=2(2)(6!)=\boxed{2880}.

解法二

思路

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也可以先选坐X桌另一边的两人，分别安排两侧，再安排Y桌。

答题过程

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Choose the two other people at table X in (62)\binom62 ways. There are 2 choices of side for Rajid and Sue, 2!2! orders for them, and 2!2! orders for the other pair. Arrange the remaining four people at table Y in 4!4! ways. Hence

N=(62)×2×2!×2!×4!=2880.N=\binom62\times2\times2!\times2!\times4!=\boxed{2880}.

(c)

解法一

思路

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相邻两人作块先算全部，再减去块处于队列任一端的情况。

答题过程

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Treat Rajid and Sue as a block with 2 internal orders. This block and the other six people have 7!7! arrangements.

The block is at an end in 2(6!)2(6!) arrangements, before accounting for its 2 internal orders. Thus

N=2(7!)−2(2)(6!)=10080−2880=7200.N=2(7!)-2(2)(6!)=10080-2880=\boxed{7200}.

解法二

思路

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也可先排另外六人，把两人组成的块插入五个内部空隙，保证两人均不在端点。

答题过程

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Arrange the other six friends in 6!6! ways. There are five interior gaps into which the Rajid–Sue block may be inserted, with 2 orders within the block. Therefore

N=6!×5×2=7200.N=6!\times5\times2=\boxed{7200}.

解法三

思路

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第三条路线先从另六人中有序选择两端，内部再排两人组成的块及另外四人。

答题过程

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Choose and order the two end people from the other six friends in 6×56\times5 ways. The interior contains the Rajid–Sue block and four other people, giving 5!5! arrangements and 2 orders within the block. Hence

N=6(5)×5!×2=7200.N=6(5)\times5!\times2=\boxed{7200}.