Permutations and Combinations | 排列与组合

📚 Permutations and Combinations | 排列与组合

Permutations and combinations are core topics in A-Level Edexcel Mathematics that underpin probability, statistics, and decision mathematics. A permutation is an arrangement of objects in a specific order, while a combination is a selection of objects where order does not matter. Mastering techniques such as factorials, the nPr and nCr formulas, and handling repeated items is essential for success in Paper 1 and applied modules.

排列与组合是 A-Level Edexcel 数学的核心课题，为概率、统计与决策数学打下基础。排列是指按特定顺序排列对象，而组合是指从对象中选取若干，不考虑顺序。掌握阶乘、nPr 与 nCr 公式以及处理重复元素等技巧，对在 Paper 1 与应用模块中取得好成绩至关重要。

1. The Factorial Function | 阶乘函数

The factorial of a positive integer n, written n!, is the product of all positive integers from 1 to n. By definition, 0! = 1, which ensures that formulas for permutations and combinations remain consistent when no objects are selected.

正整数 n 的阶乘记作 n!，表示从 1 到 n 的所有正整数的乘积。按照定义，0! = 1，这样当不选取任何对象时，排列与组合的公式仍然保持一致。

n! = n × (n − 1) × (n − 2) × … × 2 × 1

5! = 5 × 4 × 3 × 2 × 1 = 120
4! = 24, 3! = 6, 2! = 2, 1! = 1, 0! = 1

Factorials grow extremely quickly. For example, 10! = 3,628,800, and 20! is approximately 2.43 × 10¹⁸. Calculators have a factorial button, but understanding the underlying multiplication is important for simplifying algebraic expressions such as n! ÷ (n − 2)!.

阶乘增长极快。例如，10! = 3,628,800，而 20! 约为 2.43 × 10¹⁸。计算器上有阶乘键，但理解背后的乘法运算对化简 n! ÷ (n − 2)! 等代数表达式非常重要。

2. The Multiplication Principle | 乘法原理

If one operation can be performed in m ways and a second independent operation can be performed in n ways, then the two operations together can be performed in m × n ways. This principle extends to any number of operations and is the foundation for all permutation and combination counting.

如果一项操作有 m 种完成方式，另一项独立操作有 n 种完成方式，那么两项操作依次完成共有 m × n 种方式。该原理可推广到任意多项操作，是所有排列与组合计数的基础。

Choosing a shirt from 4 options and trousers from 3 options gives 4 × 3 = 12 outfits.
Rolling a die and flipping a coin gives 6 × 2 = 12 outcomes.
Selecting a starter from 5 dishes, a main from 7 dishes, and a dessert from 3 dishes gives 5 × 7 × 3 = 105 possible meals.

When applying the multiplication principle, ensure that choices at each stage are independent. If a choice affects the number of options available later, you must adjust the counts accordingly, which leads naturally to permutations.

应用乘法原理时，要确保每一步的选择相互独立。如果某个选择会影响后续可选的数量，就必须相应调整计数，这也自然引出了排列的概念。

3. Permutations of Distinct Objects | 不同对象的排列

A permutation arranges objects in order. The number of ways to arrange n distinct objects in a line is n!. This arises because there are n choices for the first position, (n − 1) for the second, and so on, down to 1 choice for the last position.

排列是将对象按顺序排成一列。将 n 个不同对象排成一行的排列数为 n!。这是因为第一个位置有 n 种选择，第二个位置有 (n − 1) 种选择，依此类推，直到最后一个位置只剩 1 种选择。

Number of arrangements of n distinct objects = n!

Arranging 3 books on a shelf: 3! = 6 arrangements.
Arranging 5 people in a queue: 5! = 120 arrangements.
Arranging the letters A, B, C, D: 4! = 24 arrangements.

This formula assumes every object is used exactly once and all objects are distinguishable. If some objects are identical, the formula must be adjusted to avoid overcounting identical arrangements.

该公式假设每个对象恰好使用一次，且所有对象均可区分。如果某些对象完全相同，则必须调整公式，以避免重复计算相同的排列。

4. Permutations of r Objects from n | 从 n 个对象中选取 r 个的排列

When selecting and arranging r objects from a set of n distinct objects, the number of permutations is denoted by ⁿPᵣ or P(n, r). The formula is n! ÷ (n − r)!, because the first position has n choices, the second has (n − 1), continuing until the r-th position has (n − r + 1) choices.

从 n 个不同对象中选取 r 个并进行排列，排列数记作 ⁿPᵣ 或 P(n, r)。公式为 n! ÷ (n − r)!，因为第一个位置有 n 种选择，第二个位置有 (n − 1) 种选择，一直到第 r 个位置有 (n − r + 1) 种选择。

ⁿPᵣ = n! ÷ (n − r)! = n × (n − 1) × (n − 2) × … × (n − r + 1)

Arranging 3 letters from 5 distinct letters: ⁵P₃ = 5! ÷ 2! = 5 × 4 × 3 = 60.
Assigning gold, silver, and bronze medals to 8 runners: ⁸P₃ = 8 × 7 × 6 = 336.
Creating 2-digit numbers from digits 1, 2, 3, 4 without repetition: ⁴P₂ = 4 × 3 = 12.

When r = n, the formula reduces to n! ÷ 0! = n!, which matches the full permutation case. When r = 0, the formula gives ⁿP₀ = n! ÷ n! = 1, representing the single empty arrangement.

当 r = n 时，公式化简为 n! ÷ 0! = n!，与全排列情形一致。当 r = 0 时，公式给出 ⁿP₀ = n! ÷ n! = 1，表示唯一的空排列。

5. Combinations | 组合

A combination is a selection of objects where order does not matter. The number of ways to choose r objects from n distinct objects is denoted by ⁿCᵣ or C(n, r), read as ‘n choose r’. The formula is n! ÷ [r!(n − r)!], which is the permutation formula divided by r! to remove ordering.

组合是从对象中选取若干，顺序无关紧要。从 n 个不同对象中选取 r 个的组合数记作 ⁿCᵣ 或 C(n, r)，读作 “n 选 r”。公式为 n! ÷ [r!(n − r)!]，即排列公式除以 r! 以消除顺序因素。

ⁿCᵣ = n! ÷ [r!(n − r)!]

Choosing 3 students from a group of 10: ¹⁰C₃ = 10! ÷ (3! × 7!) = 120.
Selecting 4 cards from a deck of 52: ⁵²C₄ = 270,725.
Picking 2 fruits from 5 different fruits: ⁵C₂ = 5! ÷ (2! × 3!) = 10.

Combinations appear in the binomial expansion as binomial coefficients, in probability as the number of favourable outcomes, and in numerous real-world selection problems. The symmetry property ⁿCᵣ = ⁿCₙ₋ᵣ is often useful for simplifying calculations.

组合在二项式展开中表现为二项式系数，在概率中表示有利结果的数量，并在众多实际选取问题中出现。对称性质 ⁿCᵣ = ⁿCₙ₋ᵣ 常常可用来简化计算。

6. Key Differences Between Permutations and Combinations | 排列与组合的关键区别

The essential question when deciding between permutations and combinations is: does the order matter? If rearranging the selected items creates a different outcome, use permutations. If the same selected items are considered identical regardless of order, use combinations.

在排列与组合之间做出选择时，关键问题是：顺序是否重要？如果重新排列所选项目会产生不同的结果，则使用排列。如果所选项目无论顺序如何都视为相同，则使用组合。

Feature	Permutation (ⁿPᵣ)	Combination (ⁿCᵣ)
Order matters?	Yes	No
Formula	n! ÷ (n − r)!	n! ÷ [r!(n − r)!]
Example	Ranking top 3 of 10	Choosing 3 of 10

Choosing a committee of 3 from 10 people: combination (order does not matter).
Choosing a president, secretary, and treasurer from 10 people: permutation (roles are distinct).
Selecting 5 lottery numbers from 50: combination (order of draw is irrelevant).

In exam questions, keywords such as ‘arrange’, ‘order’, ‘sequence’, or ‘line up’ suggest permutations, while ‘select’, ‘choose’, ‘committee’, or ‘group’ suggest combinations. Reading the question carefully is crucial.

在考试题目中，“arrange”（排列）、“order”（顺序）、“sequence”（序列）或 “line up”（排队）等关键词提示使用排列；而 “select”（选取）、“choose”（选择）、“committee”（委员会）或 “group”（小组）则提示使用组合。仔细审题至关重要。

7. Permutations with Repeated Objects | 含重复对象的排列

When arranging objects where some are identical, the total number of distinct arrangements is n! divided by the factorial of the count of each group of identical objects. This corrects for overcounting because swapping identical objects does not produce a new arrangement.

当排列含有相同元素的对象时，不同的排列总数为 n! 除以每组相同元素数量的阶乘。这样可以纠正重复计数，因为交换相同元素不会产生新的排列。

Number of arrangements = n! ÷ (p₁! × p₂! × … × pₖ!)

Arranging the letters in ‘MISSISSIPPI’: 11! ÷ (4! × 4! × 2! × 1!) = 34,650.
Arranging the letters in ‘MATHEMATICS’: 11! ÷ (2! × 2! × 2!) = 4,989,600.
Arranging the letters in ‘AAABBC’: 6! ÷ (3! × 2! × 1!) = 60.

This formula is a direct application of the multiplication principle combined with division to eliminate equivalent arrangements. It is frequently tested in Edexcel Paper 1 and in probability questions involving word arrangements.

该公式是乘法原理与除法相结合的直接应用，用于消除等价的排列。Edexcel Paper 1 以及涉及单词排列的概率题中经常考查这一内容。

8. Permutations with Restrictions | 带限制条件的排列

Many exam problems impose restrictions such as two objects must be adjacent, two objects must be separated, or certain objects must occupy specific positions. The key strategy is to treat grouped objects as a single unit or to arrange the unrestricted objects first and then insert the restricted ones.

许多考试题目会附加限制条件，例如两个对象必须相邻、两个对象必须分开，或某些对象必须占据特定位置。关键策略是将成组的对象视为一个整体，或先排列不受限制的对象，再插入受限制的对象。

Adjacent: Treat the group as one block. Arrange 5 books where 2 must stay together: 4! × 2! = 48.
Separated: Arrange the other objects first, then place the separated objects in the gaps. Arrange 3 boys and 2 girls such that no two girls sit together: 3! × ⁴P₂ = 6 × 12 = 72.
Fixed position: Fix one object, then arrange the rest. Arrange 4 people with one at a specific end: 1 × 3! = 6.

When multiple restrictions are combined, apply them sequentially and carefully count the available choices at each step. Drawing a diagram or listing positions can help avoid mistakes.

当多种限制条件同时出现时，应依次处理并仔细计算每一步可用的选择数量。画图或列出位置有助于避免错误。

9. Combinations with Restrictions | 带限制条件的组合

Combination problems with restrictions typically involve selecting a group that must include or exclude certain items. If the group must include specific items, subtract them from the total and reduce r accordingly. If items must be excluded, remove them from the available pool entirely.

带限制条件的组合问题通常涉及必须包含或排除某些项目的选取。如果小组必须包含特定项目，则从总数中减去这些项目并相应减少 r。如果必须排除某些项目，则将它们从可选池中完全移除。

Selecting 5 from 12 where 2 specific people must be included: ¹⁰C₃ = 120 (choose remaining 3 from 10).
Selecting 4 from 15 where 3 specific items must be excluded: ¹²C₄ = 495 (choose from remaining 12).
Selecting 3 from 8 where at least 1 woman must be selected from 5 women and 3 men: total ⁸C₃ − ³C₃ = 56 − 1 = 55.

The ‘at least’ or ‘at most’ conditions often require subtracting unwanted cases from the total, or summing several mutually exclusive cases. This connects directly to probability calculations in applied modules.

“至少”或“至多”条件通常需要从总数中减去不希望出现的情况，或对若干个互斥情形求和。这与应用模块中的概率计算直接相关。

10. Applications in Probability | 在概率中的应用

Permutations and combinations are fundamental tools for calculating probabilities in finite sample spaces. The probability of an event is the number of favourable outcomes divided by the total number of possible outcomes, where both counts are obtained using combinatorial methods.

排列与组合是计算有限样本空间中概率的基本工具。事件的概率等于有利结果的数量除以所有可能结果的总数，这两个数量都通过组合方法求得。

P(A) = (Number of favourable outcomes) ÷ (Total number of possible outcomes)

Probability of drawing 2 aces from a deck: ⁴C₂ ÷ ⁵²C₂ = 6 ÷ 1326 = 1/221.
Probability that 3 students selected from 10 include 2 particular friends: ⁸C₁ ÷ ¹⁰C₃ = 8 ÷ 120 = 1/15.
Probability of being dealt a flush (5 cards of one suit): 4 × ¹³C₅ ÷ ⁵²C₅ ≈ 0.00198.

In Edexcel applied papers, combinatorial probability questions often require a clear breakdown of favourable cases using combinations, especially for card, committee, and lottery scenarios.

在 Edexcel 应用试卷中，组合概率题通常要求使用组合方法清晰分解有利情形，尤其是在扑克牌、委员会和彩票场景中。

11. The Binomial Connection | 与二项式展开的联系

The binomial coefficient ⁿCᵣ is exactly the coefficient of xʳ in the expansion of (1 + x)ⁿ, or of xⁿ⁻ʳ yʳ in (x + y)ⁿ. This deep connection means that combinatorial reasoning and algebraic expansion are two views of the same underlying structure.

二项式系数 ⁿCᵣ 正是 (1 + x)ⁿ 展开中 xʳ 的系数，也是 (x + y)ⁿ 中 xⁿ⁻ʳ yʳ 的系数。这种深层联系意味着组合推理与代数展开是同一底层结构的两种视角。

(x + y)ⁿ = Σ (ⁿCᵣ) xⁿ⁻ʳ yʳ, for r = 0 to n

(x + y)³ = ³C₀ x³ + ³C₁ x²y + ³C₂ xy² + ³C₃ y³ = x³ + 3x²y + 3xy² + y³.
(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵.
The coefficient of x² in (1 + x)⁸ is ⁸C₂ = 28.

Understanding this relationship helps solve binomial expansion questions efficiently without writing Pascal’s triangle, and it provides a powerful bridge between pure mathematics and combinatorial reasoning.

理解这一关系有助于高效解决二项式展开问题，无需写出完整的帕斯卡三角。它还在纯数学与组合推理之间架起了一座强有力的桥梁。

12. Calculator Use and Exam Tips | 计算器使用与考试技巧

Edexcel A-Level calculators have dedicated buttons for nPr and nCr, usually accessed through the probability menu. Entering the value of n first, then selecting nPr or nCr, then entering r gives the answer directly. However, you must still show working in exams to earn method marks.

Edexcel A-Level 计算器设有 nPr 和 nCr 专用键，通常通过概率菜单进入。先输入 n 值，再选择 nPr 或 nCr，然后输入 r，即可直接得到答案。但在考试中仍须写出过程才能获得方法分。

Always write the formula before substituting values: ⁿCᵣ = n! ÷ [r!(n − r)!].
Check whether order matters by re-reading the question stem carefully.
For ‘at least’ questions, consider using the complement: P(A) = 1 − P(not A).
Verify small cases by listing possibilities when n is small.
Ensure final answers are in the requested form, such as a fraction or decimal to 3 significant figures.

Common mistakes include using permutations when combinations are required, forgetting to divide by repeated factorials, and miscounting gaps in separation problems. Regular practice with past paper questions will build confidence and speed.

常见错误包括在应使用组合时误用了排列、忘记除以重复元素的阶乘，以及间隔问题中数错空档。通过历年真题的常规练习，可以建立信心并提高速度。

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Permutations and Combinations | 排列与组合