📚 Changing Places: Permutations and Combinations in Edexcel A-Level Maths | 换位：Edexcel A-Level数学中的排列与组合
From shuffling cards to seating guests, “changing places” is about rearranging objects. In Edexcel A-Level Mathematics, the tools for counting these rearrangements are permutations and combinations. This article develops the key ideas step by step and shows how to avoid common pitfalls.
从洗牌到安排客人座位，“换位”本质上是重新排列对象。在Edexcel A-Level数学中，计算这些重排方式的工具是排列与组合。本文将逐步讲解核心概念，并展示如何避开常见陷阱。
1. The Multiplication Principle | 乘法原理
The multiplication principle states that if a task can be split into independent stages, the total number of outcomes is the product of the number of choices at each stage. For example, if you can choose one of 5 starters and one of 6 main courses, there are 5 × 6 = 30 possible two-course meals.
乘法原理指出，如果一项任务可以拆分为独立的阶段，那么总的结果数量等于每个阶段选择数量的乘积。例如，如果你可以选择5种开胃菜之一和6种主菜之一，那么就有 5 × 6 = 30 种可能的两道菜搭配。
This principle is crucial when we think about changing places: each position in an arrangement can be treated as a stage, and we multiply the number of choices available for each position.
这一原理在思考换位问题时至关重要：排列中的每个位置都可以看作一个阶段，我们将每个位置可用的选择数相乘。
Total outcomes = m₁ × m₂ × … × mₖ
- Example: choosing a sandwich (4 types) and a drink (3 types) gives 4 × 3 = 12 combinations. / 例子：选三明治（4种）和饮料（3种）有 4 × 3 = 12 种搭配。
- Example: selecting a two-digit code from digits 1, 2, 3, 4 without repetition gives 4 × 3 = 12 possible codes. / 例子：从数字1、2、3、4中无重复地选一个两位数编码，有 4 × 3 = 12 种可能。
2. Factorials and Basic Permutations | 阶乘与基本排列
The number of ways to arrange n distinct objects in a line is n factorial, written n!. For instance, 3! = 3 × 2 × 1 = 6, so three people can stand in a queue in six different orders.
将 n 个不同对象排列成一行的方法数是 n 的阶乘，记作 n!。例如，3! = 3 × 2 × 1 = 6，因此三个人排队可以有六种不同的顺序。
n! = n × (n − 1) × (n − 2) × … × 2 × 1
By convention, 0! = 1, which is useful in probability formulas and ensures that the permutation formula works for choosing all objects.
根据惯例，0! = 1，这在概率公式中很有用，并确保排列公式在选择全部对象时仍然成立。
- 4! = 4 × 3 × 2 × 1 = 24 / 4! = 4 × 3 × 2 × 1 = 24
- 5! = 120 / 5! = 120
- 6! = 720 / 6! = 720
3. Permutations of n Distinct Objects | n个不同对象的排列
For arranging r objects selected from n distinct objects, the number of permutations is given by n × (n − 1) × … × (n − r + 1). This is often written as nPr or P(n, r), and it counts arrangements where order matters.
对于从 n 个不同对象中选出 r 个对象进行排列，排列数由 n × (n − 1) × … × (n − r + 1) 给出。这通常记作 nPr 或 P(n, r)，它计算的是顺序重要的排列方式。
nPr = n! / (n − r)!
For example, the number of ways to arrange 3 books chosen from 5 different books on a shelf is 5P3 = 5 × 4 × 3 = 60.
例如，从5本不同的书中选出3本排列在书架上的方法数是 5P3 = 5 × 4 × 3 = 60 种。
When r = n, the formula simplifies to nPn = n!, which matches the basic permutation of all n objects.
当 r = n 时，公式简化为 nPn = n!，这与全部 n 个对象的基本排列一致。
4. Permutations with Restrictions | 带限制条件的排列
In exam questions, certain objects may be forced to stay together or be separated. If two or more objects must be together, treat them as one block. If there are n objects, reduce the number to n − 1 units, arrange these units, then arrange the objects inside the block.
在考试题中，某些对象可能被要求相邻或分开。如果两个或多个对象必须在一起，可将它们视为一个整体。如果有 n 个对象，将整体数减少到 n − 1 个单位进行排列，然后排列整体内部的对象。
For example, if A and B must sit together among 5 people, treat A and B as one block. That gives 4 units to arrange, and the block itself can be arranged in 2! ways, so the total is 4! × 2! = 48.
例如，如果 A 和 B 在5个人中必须坐在一起，将 A 和 B 视为一个整体。这样就有 4 个单位需要排列，而整体内部有 2! 种顺序，因此总数为 4! × 2! = 48。
If two objects must be separated, a common strategy is to find the total number of arrangements and subtract the arrangements where they are together. This is often safer than trying to place them directly in the gaps.
如果两个对象必须分开，常用策略是先求总排列数，再减去它们在一起的排列数。这通常比直接尝试将它们放入空隙中更安全。
5. Permutations of Non-Distinct Objects | 含有相同对象的排列
When some objects are identical, swapping identical items does not create a new arrangement. The number of distinct arrangements of n items where there are p of one type, q of another, and so on, is n! divided by the product of the factorials of the counts.
当某些对象相同时，交换相同物品不会产生新的排列。n 个物品中若有 p 个一种类型、q 个另一种类型等，则不同排列数为 n! 除以各组数量阶乘的乘积。
Number of arrangements = n! / (p! × q! × …)
This formula is essential for words or sequences with repeated letters. For instance, the number of distinct arrangements of the letters in the word ‘MAMMA’ is 5! / (3! × 2!) = 10, because there are three Ms and two As.
这个公式对于含有重复字母的单词或序列至关重要。例如，单词 ‘MAMMA’ 中字母的不同排列数是 5! / (3! × 2!) = 10，因为有三个 M 和两个 A。
6. Combinations: When Order Does Not Matter | 组合：顺序无关时的选择
When order does not matter, we use combinations. The number of ways to choose r objects from n distinct objects is written nCr or C(n, r), and it is given by dividing the number of permutations by r!.
当顺序无关紧要时，我们使用组合。从 n 个不同对象中选出 r 个对象的方法数记作 nCr 或 C(n, r)，它等于排列数除以 r!。
nCr = n! / (r! × (n − r)!)
For example, from 6 people, the number of ways to choose a committee of 2 is 6C2 = 6! / (2! × 4!) = 15. The order in which the two are selected does not matter.
例如，从6个人中选出一个2人委员会的方法数是 6C2 = 6! / (2! × 4!) = 15。这两人被选出的顺序并不重要。
7. The Relationship Between Permutations and Combinations | 排列与组合的关系
Each combination of r objects can be arranged in r! different ways. Therefore, the number of permutations is the number of combinations multiplied by r!.
每个由 r 个对象组成的组合可以排列成 r! 种不同的顺序。因此，排列数等于组合数乘以 r!。
nPr = nCr × r! or nCr = nPr / r!
This relationship helps when a question asks first to select a group and then to arrange that group. You multiply a combination by the relevant factorial.
这个关系在问题要求先选出一组、再对这组进行排列时很有用。你需要将组合数乘以相应的阶乘。
8. Circular Permutations | 圆排列
For arranging n distinct objects around a circle, rotations are considered the same. The number of circular permutations is (n − 1)!, because one object can be fixed as a reference point, and the remaining n − 1 objects are arranged linearly.
将 n 个不同对象围绕圆圈排列时，旋转被视为相同。圆排列数为 (n − 1)!，因为可以固定一个对象作为参考点，然后对其余 n − 1 个对象进行线性排列。
Circular permutations = (n − 1)!
For example, the number of ways to seat 5 people around a round table is (5 − 1)! = 4! = 24.
例如，让5个人围坐在一张圆桌旁的方法数是 (5 − 1)! = 4! = 24 种。
9. Application: Probability and Counting | 应用：概率与计数
Permutations and combinations appear in probability questions, often as the denominator or numerator of a probability fraction. The probability of a favourable outcome is found by dividing the number of favourable arrangements by the total number of arrangements.
排列与组合出现在概率问题中，通常作为概率分数的分母或分子。一个有利结果的概率等于有利排列数除以总排列数。
P(event) = favourable arrangements / total arrangements
For example, if 4 letters are randomly arranged from the letters A, B, C, D, E, the probability that the arrangement starts with A is 4P3 / 5P4 = 24 / 120 = 1/5.
例如，如果从字母 A、B、C、D、E 中随机排列4个字母
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