题目

Problem

In a group of 25 athletes, there are 8 sprinters, 5 hurdlers and 12 throwers.

(a) Find the number of different ways in which a team of 6 athletes can be selected if it consists of at least 3 sprinters, at most 2 hurdlers and at most 1 thrower.

(4)

A group of 8 athletes chosen from the group of 25 athletes consists of 1 sprinter, 3 hurdlers and 4 throwers. These 8 athletes stand in a row.

(b) How many different arrangements of the 8 athletes are there if the 3 hurdlers do not stand all together?

(3)

(c) How many different arrangements of the 8 athletes are there in which there are at least two athletes between any two hurdlers?

(3)

题目中文翻译

一组 25 名运动员中，有 8 名短跑运动员、5 名跨栏运动员和 12 名投掷运动员。

(a) 若要选出一支由 6 名运动员组成的队伍，且其中至少有 3 名短跑运动员、至多有 2 名跨栏运动员、至多有 1 名投掷运动员，求不同的选法数。

(4)

从这 25 名运动员中选出的 8 人包括 1 名短跑运动员、3 名跨栏运动员和 4 名投掷运动员。这 8 人站成一排。

(b) 若 3 名跨栏运动员不全部相邻站立，这 8 人共有多少种不同的排列？

(3)

(c) 若任意两名跨栏运动员之间至少有两名运动员，这 8 人共有多少种不同的排列？

(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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同时满足三条人数限制，按队伍的短跑、跨栏、投掷人数列出所有六种构成。每一种用组合数计数，再相加。

答题过程

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Let S, H and T denote sprinters, hurdlers and throwers. The permitted team compositions are:

S	H	T	Number of selections
$6$	$0$	$0$	$\binom86=28$
$5$	$1$	$0$	$\binom85\binom51=280$
$5$	$0$	$1$	$\binom85\binom{12}{1}=672$
$4$	$2$	$0$	$\binom84\binom52=700$
$4$	$1$	$1$	$\binom84\binom51\binom{12}{1}=4200$
$3$	$2$	$1$	$\binom83\binom52\binom{12}{1}=6720$

Each row has at least 3 sprinters, at most 2 hurdlers and at most 1 thrower, and the rows exhaust the possibilities. Therefore,

28+280+672+700+4200+6720=\boxed{12600}.

(b)

解法一

思路

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8人各不相同，先算全部排列，再减去三个跨栏选手全部相连的排列。作为一个块后，还要乘块内三人的排列数。

答题过程

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All eight athletes are distinct, giving $8!$ arrangements.

If the three hurdlers are all together, treat them as one block. Together with the five other athletes there are six objects, arranged in $6!$ ways, and the hurdlers can be ordered internally in $3!$ ways.

Thus the required number is

8!-6!\,3!=40320-4320=\boxed{36000}.

解法二

思路

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另一条路线先排五位非跨栏选手，形成六个空隙。三个跨栏选手要么各占不同空隙，要么两人一组、另一人独立，不包括三人同一空隙。

答题过程

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Arrange the five non-hurdlers in $5!$ ways, creating six gaps including the end gaps.

All three hurdlers separate: choose three gaps and arrange the hurdlers, giving $\binom63\,3!$ possibilities.
Exactly two hurdlers together: choose the gap for the pair and a different gap for the single hurdler, then arrange the three hurdlers, giving $6\times5\times3!$ possibilities.

Thus

\begin{align*} N=&\,5!\,3!\biggl(\binom63+6\times5\biggr)\\[4mm] =&\,14400+21600\\[4mm] =&\,\boxed{36000}. \end{align*}

(c)

解法一

思路

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三个跨栏选手之间的两个内部间隔各至少需要两位其他选手，先用掉四人；剩下一位可以放在前端、后端或两个内部间隔之一，共四种位置模式。

答题过程

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Write H for a hurdler and N for a non-hurdler. There are five non-hurdlers.

The two internal gaps between consecutive hurdlers must each contain at least two non-hurdlers. After placing these four, the one remaining non-hurdler can be added to either end gap or either internal gap. Thus there are four possible patterns.

The five distinct non-hurdlers can be assigned to their positions in $5!$ ways, and the three distinct hurdlers in $3!$ ways. Hence

N=4\times5!\times3!=\boxed{2880}.

CIE 9709 2025 November Paper 55 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程

(c)

解法一

思路

答题过程