📚 The Selection of Members | 成员的选择
In Edexcel A Level Mathematics, questions on selecting members from a group test your understanding of permutations and combinations. This article explains the key ideas, formulas, and common restrictions step by step.
在爱德思 A Level 数学中,关于从一组中选择成员的问题考查你对排列与组合的理解。本文逐步解释核心概念、公式和常见限制条件。
1. The Fundamental Counting Principle | 基本计数原理
If there are m ways to perform task A and n ways to perform task B independently, then there are m × n ways to perform both tasks in order. For example, choosing a chair from 5 candidates and a secretary from 4 remaining candidates gives 5 × 4 = 20 possible pairs.
若任务 A 有 m 种完成方式,任务 B 有 n 种独立完成方式,则按顺序完成两个任务共有 m × n 种方式。例如,从 5 名候选人中选主席、再从剩余 4 人中选秘书,共有 5 × 4 = 20 种选择。
This principle extends to any number of stages: multiply the number of choices at each stage. It is the foundation of both permutations and combinations.
该原理可扩展到任意多个阶段:将每一阶段的选择数相乘。它是排列与组合的基础。
2. Factorial Notation | 阶乘记号
For a positive integer n, n! = n × (n – 1) × … × 2 × 1. By convention, 0! = 1. Factorials count the number of ways to arrange n distinct objects in a line.
对于正整数 n,n! = n × (n – 1) × … × 2 × 1。按约定,0! = 1。阶乘用于计算 n 个不同对象排成一行的方法数。
You should be comfortable simplifying expressions such as n! / (n – 2)! = n(n – 1), because many exam questions require cancellation before calculation.
你应熟练化简表达式,例如 n! / (n – 2)! = n(n – 1),因为许多考题要求先约分再计算。
3. Introducing Permutations | 排列介绍
A permutation is an arrangement of objects where order matters. When selecting r members from n distinct members and arranging them, the number of permutations is P(n, r) = n! / (n – r)!.
排列是考虑顺序的对象安排。从 n 个不同成员中选出 r 个并安排顺序,排列数为 P(n, r) = n! / (n – r)!。
For example, the number of ways to choose a chair, vice-chair, and secretary from 8 members is P(8, 3) = 8! / 5! = 8 × 7 × 6 = 336.
例如,从 8 名成员中选出主席、副主席和秘书的方法数为 P(8, 3) = 8! / 5! = 8 × 7 × 6 = 336。
4. Introducing Combinations | 组合介绍
A combination is a selection of objects where order does not matter. The number of ways to select r members from n distinct members is C(n, r) = n! / (r! (n – r)!).
组合是不考虑顺序的对象选择。从 n 个不同成员中选出 r 个的方法数为 C(n, r) = n! / (r! (n – r)!)。
For example, choosing a 3-person committee from 8 members can be done in C(8, 3) = 56 ways, because the order of the three chosen members is irrelevant.
例如,从 8 名成员中选出 3 人委员会有 C(8, 3) = 56 种方式,因为所选 3 人的顺序无关紧要。
5. The Combination Formula nCr | 组合公式 nCr
The formula C(n, r) = n! / (r! (n – r)!) is derived by taking the permutation count P(n, r) = n! / (n – r)! and dividing by r! to remove the ordering of the chosen r members.
公式 C(n, r) = n! / (r! (n – r)!) 的推导方法是:先取排列数 P(n, r) = n! / (n – r)!,再除以 r! 以消除所选 r 个成员的顺序。
Useful identities include C(n, r) = C(n, n – r), and C(n, 0) = C(n, n) = 1. These identities can save time in exams.
常用恒等式包括 C(n, r) = C(n, n – r),以及 C(n, 0) = C(n, n) = 1。这些恒等式可帮助你在考试中节省时间。
Remember that 0 ≤ r ≤ n must always hold; if r is outside this range, C(n, r) is defined as 0 in most contexts.
记住 0 ≤ r ≤ n 必须始终成立;如果 r 超出此范围,大多数情况下 C(n, r) 被定义为 0。
6. Selecting from Distinct vs Identical Members | 从不同成员与相同成员中选择
In most Edexcel questions, members are distinct individuals, so combinations and permutations apply directly. Be alert for wording like ‘identical’ or ‘indistinguishable’, which changes the counting method.
在大多数爱德思考题中,成员是不同的个体,因此可直接使用组合与排列。注意题目中’相同’或’不可区分’等措辞,这些会改变计数方法。
If some members are identical, you may need to divide by the factorial of the number of identical items, or list possible cases separately. However, A Level questions rarely go beyond simple distinct-member selections.
如果某些成员相同,你可能需要除以相同物品数量的阶乘,或分别列出可能情况。不过 A Level 题目很少超出简单的不同成员选择。
7. Restrictions: Must Include or Exclude | 限制条件:必须包含或排除
When a question says a specific member must be included, first select that member (1 way), then choose the remaining r – 1 from the other n – 1 members: C(n – 1, r – 1).
当题目说明某特定成员必须被选入时,先选择该成员(1 种方式),再从其余 n – 1 名成员中选 r – 1 名:C(n – 1, r – 1)。
If a specific member must be excluded, simply reduce the pool to n – 1 members and select r from them: C(n – 1, r). Treating restrictions first avoids double counting.
如果某特定成员必须被排除,只需将可选范围缩小为 n – 1 名成员,从中选 r 名:C(n – 1, r)。先处理限制条件可避免重复计数。
8. Multi-stage Selections | 多阶段选择
Many exam problems require selecting members from different groups, such as 3 boys from 10 boys and 2 girls from 8 girls. Use the multiplication principle: C(10, 3) × C(8, 2).
许多考题要求从不同组中选择成员,例如从 10 名男生中选 3 名、从 8 名女生中选 2 名。使用乘法原理:C(10, 3) × C(8, 2)。
Be careful: if the problem only specifies the total number of members without group restrictions, you should select from the combined pool, not multiply separate group counts.
注意:如果题目只指定总人数而没有分组限制,则应从合并后的人群中选择,而不是将各组计数相乘。
Always check whether stages are independent; if one choice affects the next, adjust the numbers accordingly.
始终检查各阶段是否独立;如果前一个选择影响后一个选择,要相应地调整数字。
9. Using Selection in Probability | 在概率中应用选择
To find a probability involving a selection, compute the number of favourable outcomes and divide by the total number of possible selections. Both are usually combinations.
要求涉及选择的概率时,计算有利结果数并除以所有可能选择的总数。两者通常都用组合数表示。
For example, if 3 members are chosen at random from 10, the probability that a particular member A is included equals C(1,1) × C(9,2) / C(10,3) = 36 / 120 = 0.3.
例如,从 10 人中随机选 3 人,特定成员 A 被选中的概率为 C(1,1) × C(9,2) / C(10,3) = 36 / 120 = 0.3。
Always ensure the numerator and denominator are counted under the same rules regarding order and restrictions.
始终确保分子与分母在顺序和限制条件方面使用相同的计数规则。
10. Common Exam Mistakes | 常见考试错误
The most common mistake is using permutations when order does not matter, or combinations when order does. Always ask: does swapping two selected members create a different outcome?
最常见的错误是在顺序无关时使用排列,或在顺序有关时使用组合。始终问自己:交换两个所选成员是否会得到不同结果?
Another frequent error is forgetting to subtract excluded members before counting, or multiplying when you should add. Double check whether cases are mutually exclusive or independent.
另一个常见错误是在计数前忘记减去被排除的成员,或在应该相加时错误相乘。反复检查各情况是互斥的还是独立的。
Also watch out for arithmetic slips with factorials; simplify the formula before using a calculator.
还要注意阶乘的算术失误;在使用计算器之前先化简公式。
11. Worked Example: Committee Selection | 例题:委员会选择
Worked example: From 12 people, a committee of 5 is to be chosen. If one particular person must be on the committee, how many committees are possible?
例题:从 12 人中选出一个 5 人委员会。如果某特定人员必须进入委员会,共有多少种可能的委员会?
Solution: First include the required person: 1 way. Choose the remaining 4 from the other 11 people: C(11, 4) = 330. Thus there are 1 × 330 = 330 possible committees.
解:首先纳入该特定人员:1 种方式。从其余 11 人中选 4
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