📚 Mastering Combinations: nCr and Counting Principles | 组合数学：nCr 与计数原理
In A-level mathematics, combinations are a core tool for counting selections where order does not matter. From choosing committee members to calculating lottery probabilities, the idea of 'n choose r' appears across probability, algebra and statistics. This article explains the definition, formula, properties and common exam applications of combinations, with worked examples and tips to avoid common errors.
在 A-level 数学中，组合是计算顺序无关的选择数量的核心工具。从挑选委员会成员到计算彩票概率，“n 选 r”的概念广泛出现在概率、代数和统计中。本文讲解组合的定义、公式、性质以及常见考试应用，并提供例题和避免常见错误的技巧。
1. What is a Combination? | 什么是组合？
A combination is a selection of items from a larger set where the arrangement or order of the selected items does not matter. For example, choosing the letters A, B and C from {A, B, C, D, E} is the same combination as choosing B, C, A. The subset {A, B, C} is counted once, regardless of internal order.
组合是从一个较大集合中选取若干项的方式，所选项目的排列顺序无关紧要。例如，从 {A, B, C, D, E} 中选出字母 A、B、C，与选择 B、C、A 是同一个组合。子集 {A, B, C} 只计一次，无论内部顺序如何。
In practical problems, words like 'select', 'choose', 'form a group' or 'pick a team' usually indicate that combinations should be used. Words like 'arrange', 'line up' or 'order' suggest permutations instead.
在实际问题中，“选取”、“选择”、“组成小组”或“挑选团队”等词语通常表明应使用组合。而“安排”、“排队”或“排序”等词语则提示应使用排列。
2. Definition and Notation | 定义与记号
denoted by C(n, r), nCr, or the binomial coefficient (n choose r). The symbols are read as 'n choose r' r is the number of items selected. Both n and r are non-negative integers with 0 ≤ r ≤ n.
从 n 个不同对象中选出 r 个对象的方法数记作 C(n, r)、nCr 或二项式系数 (n choose r)。这些记号读作“n 选 r”。在这个记号中，n 是可用项目总数，r 是所选项目数。n 和 r 都是非负整数，满足 0 ≤ r ≤ n。
C(n, r) = n! / [r!(n - r)!]
The exclamation mark denotes factorial: n! = n × (n - 1) × ... × 2 × 1, with 0! defined as 1. For instance, 5! = 5 × 4 × 3 × 2 × 1 = 120, and 0! = 1 by definition.
感叹号表示阶乘：n! = n × (n - 1) × ... × 2 × 1，并定义 0! = 1。例如，5! = 5 × 4 × 3 × 2 × 1 = 120，而 0! = 1 是规定。
Some textbooks also use the vector-style notation ⁿCᵣ or the matrix form. They all refer to exactly the same counting value, so you should recognise all three when reading exam questions.
一些教材还使用向量式记号 ⁿCᵣ 或矩阵形式。它们都指完全相同的计数值，因此在阅读考题时应能识别这三种写法。
3. Why Order Does Not Matter | 为什么顺序无关
Combinations count selections as sets, not sequences. If a question asks 'how many ways can three students be selected from ten?', the group {Alice, Ben, Cara} is identical to {Ben, Cara, Alice}, so it is one combination.
组合把选择结果当作集合而不是序列来计数。如果题目问“从十名学生中选出三名有多少种方式？”，则 {Alice, Ben, Cara} 与 {Ben, Cara, Alice} 完全相同，因此只算一个组合。
If order did matter, we would divide by r! because each selected set of r items can be arranged in r! different orders. This gives the relationship nPr = nCr × r! or equivalently C(n, r) = nPr / r!.
如果顺序重要，我们需要除以 r!，因为每个选出的 r 项集合可以按 r! 种不同顺序排列。由此得到关系 nPr = nCr × r!，等价地 C(n, r) = nPr / r!。
nPr = n! / (n - r)! and C(n, r) = nPr / r!
This relationship is extremely useful in mixed questions: first count ordered arrangements with nPr, then remove the internal ordering by dividing by r!.
这个关系在混合题型中非常有用：先用 nPr 计算有序安排，然后除以 r! 去除内部顺序。
4. Key Difference: Permutations vs Combinations | 排列与组合的关键区别
Permutations count ordered arrangements, while combinations count unordered selections. The word 'arrange' usually signals permutation; 'select', 'choose' or 'form a group' often signals combination.
排列计数有序的安排，组合计数无序的选择。“安排”一词通常提示排列；“选取”、“选择”或“组成小组”通常提示组合。
- Permutation: order matters — electing a president, secretary and treasurer from 8 people gives 8P3. 排列：顺序重要——从8人中选举主席、秘书和财务，有 8P3 种。
- Combination: order does not matter — choosing 3 representatives from 8 people gives 8C3. 组合：顺序无关——从8人中选3名代表，有 8C3 种。
A quick check is to ask: if I swap two selected items, do I get a new outcome? If yes, use permutations; if no, use combinations. This simple test prevents the most common counting error in exams.
一个快速的检验方法是问：交换两个已选项后，是否得到新的结果？如果是，用排列；如果不是，用组合。这个简单的检验可以防止考试中最常见的计数错误。
5. Properties of Combinations | 组合的性质
C(n, n - r). Choosing r items to include is equivalent to choosing n - r items to exclude.
组合具有有用的对称性和恒等性质。一个关键性质是 C(n, r) = C(n, n - r)。选择 r 项包含等同于选择 n - r 项排除。
C(n, r) = C(n, n - r)
Another identity is Pascal's relation: C(n, r) = C(n - 1, r - 1) + C(n - 1, r). This underlies Pascal's triangle and binomial expansions.
另一个恒等式是帕斯卡关系：C(n, r) = C(n - 1, r - 1) + C(n - 1, r)。它是帕斯卡三角形和二项式展开的基础。
C(n, r) = C(n - 1, r - 1) + C(n - 1, r)
= n for n ≥ 1. These values simplify many calculations and allow quick checks of algebra.
还请注意边界值：C(n, 0) = 1，C(n, n) = 1，且当 n ≥ 1 时 C(n, 1) = n。这些值可以简化许多计算，并用于快速检验代数结果。
6. Worked Example: Choosing a Committee | 例题：选择委员会
Problem: A club has 10 members. In how many ways can a committee of 4 be chosen?
问题：某俱乐部有 10 名成员。从中选出 4 人组成委员会，共有多少种方式？
Solution: Order does not matter because a committee is a group, not an ordered list. Use C(10, 4).
解答：顺序无关，因为委员会是一个小组，不是有序名单。使用 C(10, 4)。
C(10, 4) = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 210
Therefore, there are 210 possible committees.
因此，共有 210 种可能的委员会。
ecretary and treasurer, order would matter and the answer would be 10P4 = 5040, which is 4! times larger because the four roles can be rearranged within each group of four.
如果同一问题要求选出主席、副主席、秘书和财务，则顺序重要，答案将是 10P4 = 5040，这是 4! 倍，因为四个职位可以在每组四人内部重新排列。
7. Combinations with Restrictions | 带限制条件的组合
Many exam questions combine combinations with inclusion, exclusion, or grouping conditions. A common pattern is: 'The committee must include at least one woman' or 'two particular members cannot both be selected.'
许多考题将组合与包含、排除或分组条件结合。常见模式如：“委员会必须至少包含一名女性”或“某两名特定成员不能同时入选”。
Example: From 6 men and 5 women, how many 4-person committees contain exactly 2 women?
例题：从 6 名男性和 5 名女性中选出 4 人委员会，恰好包含 2 名女性的方式有多少？
Solution: Choose 2 women from 5 and 2 men from 6 independently, then multiply.
解答：从 5 名女性中选 2 人，从 6 名男性中选 2 人，独立选择后相乘。
C(5, 2) × C(6, 2) = 10 × 15 = 150
So there are 150 such committees. For 'at least' questions, break into cases and add, or use the complement method. For example, 'at least one woman' equals total committees minus committees with no women.
因此共有 150 个这样的委员会。对于“至少”类问题，可以分情况相加，或使用补集法。例如，“至少一名女性”等于委员会总数减去没有女性的委员会数。
When a restriction says that two particular people cannot both be selected, a good strategy is to subtract the number of committees containing both from the unrestricted total.
当限制条件说某两个特定人员不能同时入选时，一个有效策略是从无限制总数中减去同时包含这两人的委员会数。
8. Applications in Probability | 在概率中的应用
ly likely, probability = number of favourable combinations / total number of combinations.
组合用于计算不放回选择中的概率。如果所有结果等可能性，则概率 = 有利组合数 / 总组合数。
Example: A bag contains 8 red and 4 blue balls. If 3 balls are drawn at random without replacement, what is the probability that exactly 2 are red?
例题：一个袋子中有 8 个红球和 4 个蓝球。随机不放回抽取 3 个球，恰好 2 个红球的概率是多少？
Solution: Total ways = C(12, 3). Favourable ways = C(8, 2) × C(4, 1).
解答：总方式 = C(12, 3)。有利方式 = C(8, 2) × C(4, 1)。
P = [C(8, 2) × C(4, 1)] / C(12, 3) = (28 × 4) / 220 = 112 / 220 = 28 / 55
9. This combinatorial probability method is standard in A-level statistics and often appears
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