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CIE 9709 2025 November Paper 53 Q3

A Level / CIE / S1

CIE 9709 2025 Nov Paper 53 Paper · Question 3

题目

Problem

(a) Find the number of different arrangements of the 9 letters in the word DAFFODILS in which there is a D at each end and the two Fs are not next to each other.

(3)

(b) Find the probability that a randomly chosen arrangement of the 9 letters in the word DAFFODILS has exactly 4 letters between the two Ds.

(3)
题目中文翻译

(a) 求将单词 DAFFODILS 的 9 个字母排列成两端各有一个 D,且两个 F 不相邻的不同排列数。

(3)

(b) 随机选取单词 DAFFODILS 的 9 个字母的一种排列,求两个 D 之间恰有 4 个字母的概率。

(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

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固定首尾的D以后,内部7个字母中两个F相同。先算全部排列,再减去FF相邻的排列。

答题过程

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Fix a D at each end. The remaining seven letters are A, F, F, O, I, L, S, giving 7!2!\frac{7!}{2!} arrangements.

If FF is treated as one block, there are six distinct objects, giving 6!6! arrangements. Hence

7!2!−6!=2520−720=1800.\frac{7!}{2!}-6!=2520-720=\boxed{1800}.

解法二

思路

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也可以先排五个不同字母A、O、I、L、S,再把两个F放在六个不同空隙中。

答题过程

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Arrange A, O, I, L, S in 5!5! ways between the fixed Ds. These letters create six gaps, all between the two Ds.

Choose two distinct gaps and place one F in each. Thus the required number is

5!(62)=120×15=1800.5!\binom62=120\times15=\boxed{1800}.

(b)

解法一

思路

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两个D之间恰有4个字母,位置相差5。左D有4种位置,剩下7格含两个相同的F;最后除以全部不同排列数。

答题过程

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The possible position pairs for the Ds are (1,6),(2,7),(3,8),(4,9)(1,6),(2,7),(3,8),(4,9).

For each pair, the other seven letters have 7!2!\frac{7!}{2!} arrangements, so the number of favourable arrangements is

47!2!=10080.4\frac{7!}{2!}=10080.

The total number of different arrangements is 9!2! 2!=90720\frac{9!}{2!\,2!}=90720. Therefore,

P=1008090720=19.\mathrm{P}=\frac{10080}{90720}=\boxed{\frac19}.

解法二

思路

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另一条路线按夹在D之间的F数分类,将D及中间四字母合成块。0、1、2个F的三种情况需要分别处理相同F的排列。

答题过程

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The five letters other than D and F are A, O, I, L, S.

Treat the string D____D as one block. Count according to the number of Fs inside it:

  • No F inside: choose and order four of the five distinct letters, then arrange the block, the remaining distinct letter and two identical Fs:
5P4 4!2!=1440.{}_5\mathrm{P}_4\,\frac{4!}{2!}=1440.
  • One F inside: choose three of the five distinct letters and arrange them with F. The block and the three remaining distinct letters have 4!4! arrangements:
(53) 4! 4!=5760.\binom53\,4!\,4!=5760.
  • Two Fs inside: choose two of the five distinct letters, arrange the four internal letters accounting for the identical Fs, then arrange the block and the three remaining letters:
(52) 4!2! 4!=2880.\binom52\,\frac{4!}{2!}\,4!=2880.

Therefore,

P=1440+5760+28809!/(2! 2!)=19.\mathrm{P}=\frac{1440+5760+2880}{9!/(2!\,2!)} =\boxed{\frac19}.