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CIE 9709 2024 November Paper 52 Q2

A Level / CIE / S1

CIE 9709 2024 Nov Paper 52 Paper · Question 2

题目

Problem

(a) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC.

(1)

(b) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC in which there are no more than two letters between the two As.

(3)
题目中文翻译

(a) 求单词 ALGEBRAIC 中 9 个字母的不同排列数。

(1)

(b) 求单词 ALGEBRAIC 中 9 个字母的不同排列数,要求两个 A 之间不超过两个字母。

(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

9个字母中只有A重复两次。

答题过程

展开 N=9!2!=181440.N=\frac{9!}{2!}=\boxed{181440}.

(b)

解法一

思路

展开

按两A之间有0、1、2个字母分类。固定左A位置后,其余七个不同字母可任意排列。

答题过程

展开

If there are rr letters between the As, the left A can occupy 8−r8-r positions. Once both A positions are fixed, the other seven distinct letters have 7!7! arrangements.

Letters between AsNumber of arrangements
008×7!=403208\times7!=40320
117×7!=352807\times7!=35280
226×7!=302406\times7!=30240

Thus

N=(8+7+6)7!=105840.N=(8+7+6)7!=\boxed{105840}.

解法二

思路

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另一条路线先选出两A中间的有序字母,再把两A及中间字母看作一个块。

答题过程

展开

Classify by the number of letters between the As.

  • No letter: the AA block and seven other letters have 8!8! arrangements.
  • One letter: choose that letter in 7 ways, then arrange the A_A block and six others in 7!7! ways.
  • Two letters: choose and order them in 7×67\times6 ways, then arrange the A__A block and five others in 6!6! ways.

Hence

N= 8!+7(7!)+7(6)(6!)= 40320+35280+30240=105840.\begin{align*} N=&\,8!+7(7!)+7(6)(6!)\\[4mm] =&\,40320+35280+30240=\boxed{105840}. \end{align*}

解法三

思路

展开

也可以用(a)的全部排列减去两A间隔3至7个字母的情况。

答题过程

展开

For r=3,4,5,6,7r=3,4,5,6,7 letters between the As, there are (8−r)7!(8-r)7! arrangements. Therefore the required number is

N= 9!2!−(5+4+3+2+1)7!= 181440−75600=105840.\begin{align*} N=&\,\frac{9!}{2!}-(5+4+3+2+1)7!\\[4mm] =&\,181440-75600=\boxed{105840}. \end{align*}