题目

Problem

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

(2)

(b) Find the number of different arrangements of the 9 letters in the word ACTIVATED in which there are at least 5 letters between the two As.

(3)

Five letters are selected at random from the 9 letters in the word ACTIVATED.

(5)

题目中文翻译

(a) 求单词 ACTIVATED 中 9 个字母的不同排列数。

(2)

(b) 求单词 ACTIVATED 中 9 个字母的不同排列数，要求两个 A 之间至少有 5 个字母。

(3)

从单词 ACTIVATED 的 9 个字母中随机选取 5 个字母。

(5)

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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两个A、两个T分别重复。

答题过程

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N=\frac{9!}{2!\,2!}=\boxed{90720}.

(b)

解法一

思路

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两A之间5、6、7个字母，对应左A位置3、2、1种，其余七格中两个T相同。

答题过程

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If there are $r$ letters between the As, the left A has $8-r$ possible positions. The remaining seven letters have $\frac{7!}{2!}$ arrangements, since the Ts are identical.

Thus the required count is

N=(3+2+1)\frac{7!}{2!}=\boxed{15120}.

(c)

解法一

思路

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随机抽的是实际字母牌，组合样本126种。按A、T本数列T不多于A的六类。

答题过程

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There are $\binom95=126$ equally likely selections of five individual letters. Besides two As and two Ts, there are five other letters.

As	Ts	Other letters	Selections
$0$	$0$	$5$	$1$
$1$	$0$	$4$	$\binom21\binom54=10$
$1$	$1$	$3$	$\binom21\binom21\binom53=40$
$2$	$0$	$3$	$\binom53=10$
$2$	$1$	$2$	$\binom21\binom52=20$
$2$	$2$	$1$	$\binom51=5$

Hence

\mathrm{P}=\frac{1+10+40+10+20+5}{126}=\boxed{\frac{43}{63}}.

解法二

思路

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也可数T比A多的三种不合格构成，再用补集。

答题过程

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The cases with more Ts than As are $(T,A)=(1,0),(2,0),(2,1)$ . Their counts are

\begin{align*} \binom21\binom54=&\,10,\\[4mm] \binom53=&\,10,\\[4mm] \binom21\binom52=&\,20. \end{align*}

Thus

\mathrm{P}=1-\frac{10+10+20}{\binom95}=\boxed{\frac{43}{63}}.

CIE 9709 2022 November Paper 53 Q6

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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解法二

思路

答题过程