Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 March Paper 52 Q2

A Level / CIE / S1

CIE 9709 2026 March Paper 52 Paper · Question 2

题目

Problem

(a) Find the number of different arrangements of the 9 letters in the word ATTENTIVE in which the two Es are together but the three Ts are not all together.

(3)

(b) Find the number of different arrangements of the 9 letters in the word ATTENTIVE in which there are at least four letters between the two Es.

(4)
题目中文翻译

(a) 求将单词 ATTENTIVE 的 9 个字母排列成两个 E 相连,但三个 T 不全部相连的不同排列数。

(3)

(b) 求将单词 ATTENTIVE 的 9 个字母排列成两个 E 之间至少有四个字母的不同排列数。

(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

展开

先把EE合为一块,排列剩下各字母,再减去TTT也合为一块的排列。

答题过程

展开

Treat EE as a single block. The eight objects are EE, A, T, T, T, N, I and V, giving 8!3!\frac{8!}{3!} arrangements.

If the three Ts are also together, treat TTT as a block. There are then six distinct objects, giving 6!6! arrangements.

Thus the required number is

8!3!−6!=6720−720=6000.\frac{8!}{3!}-6!=6720-720=\boxed{6000}.

解法二

思路

展开

也可以先排EE块和A、N、I、V五个对象,再在六个空隙中放三个T。排除全部T在同一个空隙,只剩三个分开或两个相连的情况。

答题过程

展开

Arrange EE, A, N, I and V in 5!5! ways, creating six gaps including the end gaps.

For the Ts not all to be together:

  • three separate Ts occupy three distinct gaps: (63)=20\binom63=20 choices;
  • a pair TT and a single T occupy two different gaps: 6×5=306\times5=30 choices.

Hence

5!((63)+6×5)=120(20+30)=6000.5!\biggl(\binom63+6\times5\biggr) =120(20+30) =\boxed{6000}.

(b)

解法一

思路

展开

两个E之间可有4、5、6、7个字母。对每一种间隔,数出左E能放的位置;其余7个字母中三个T相同。

答题过程

展开

If there are rr letters between the Es, the left E can be in 8−r8-r positions.

Letters between Es, rrChoices for E positionsArrangements
444447!3!=33604\frac{7!}{3!}=3360
553337!3!=25203\frac{7!}{3!}=2520
662227!3!=16802\frac{7!}{3!}=1680
77117!3!=840\frac{7!}{3!}=840

Thus the total is

3360+2520+1680+840=8400.3360+2520+1680+840=\boxed{8400}.

解法二

思路

展开

也可以将两E和中间字母作为一个块。先暂时标记三个T,按块内字母数选取和排列,再排列块与余下字母,最后除去标签。

答题过程

展开

Temporarily label the three Ts, so the seven non-E letters are distinct.

If rr letters are between the Es, choose and order them in  7Pr\,{}_7\mathrm{P}_r ways. The block E…E and the 7−r7-r remaining letters can be arranged in (8−r)!(8-r)! ways. Removing the T labels divides by 3!3!.

For r=4,5,6,7r=4,5,6,7, the total number is

N= 7P4 4!+7P5 3!3! +7P6 2!+7P7 1!3!= 3360+2520+1680+840= 8400.\begin{align*} N=&\,\frac{{}_7\mathrm{P}_4\,4!+{}_7\mathrm{P}_5\,3!}{3!}\\[4mm] &\,\hspace{2pt}+\frac{{}_7\mathrm{P}_6\,2!+{}_7\mathrm{P}_7\,1!}{3!}\\[4mm] =&\,3360+2520+1680+840\\[4mm] =&\,\boxed{8400}. \end{align*}