The Cabinet: Permutations, Combinations and Ministerial Selections | 内阁：排列组合与部长选拔

📚 The Cabinet: Permutations, Combinations and Ministerial Selections | 内阁：排列组合与部长选拔

In Edexcel A Level Mathematics, the topic of permutations and combinations often appears in the Pure Mathematics specification under counting principles and binomial expansion. This article uses the idea of forming a government cabinet to show how these counting tools answer real-world questions such as ‘How many ways can we choose ministers?’ and ‘What is the probability of a particular team being selected?’

在 Edexcel A Level 数学中，排列与组合常出现在纯数学的计数原理和二项式展开部分。本文以内阁组阁为背景，展示如何运用这些计数工具回答现实问题，例如 “有多少种方式选出部长？” 以及 “某一特定团队被选中的概率是多少？”

By working through structured examples, you will revise factorial notation, combinations, permutations, restricted selections and probability calculations. Each section pairs an English explanation with a Chinese translation to strengthen both mathematical understanding and bilingual terminology.

通过结构化例题，你将复习阶乘表示、组合、排列、受限选择以及概率计算。每个部分都配有英文解释和中文翻译，以加深数学理解和双语术语掌握。

1. The Basic Counting Principle | 基本计数原理

The product rule for counting states that if one operation can be performed in m ways and a second independent operation can be performed in n ways, then the two operations can be performed together in m × n ways.

计数乘法原理指出：如果第一个操作有 m 种方式，第二个独立操作有 n 种方式，那么依次完成这两个操作共有 m × n 种方式。

When forming a cabinet, suppose there are 4 candidates for Prime Minister and, for each choice, 6 possible Deputy Prime Ministers. The total number of leader pairings is therefore 4 × 6 = 24.

组阁时，假设首相候选人有 4 位，而每位首相人选对应 6 位副首相候选人。那么领导组合总数就是 4 × 6 = 24。

English: Multiply the number of choices at each stage only when the choices are independent.
中文：只有在各阶段选择相互独立时，才将各阶段的选择数相乘。
English: The product rule extends to any number of stages: m × n × p × …
中文：乘法原理可推广到任意多个阶段：m × n × p × …

2. Factorial Notation and Ministerial Line-ups | 阶乘表示与部长排序

For a positive integer n, n factorial is defined as n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1. By convention, 0! = 1.

对于正整数 n，n 的阶乘定义为 n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1。约定 0! = 1。

If 5 ministers are to be arranged in a row for a photograph, the number of possible orders is 5! = 120. This is because the first position can be filled in 5 ways, the next in 4 ways, and so on.

如果 5 位部长要排成一排拍照，可能的顺序数为 5! = 120。这是因为第一个位置有 5 种选择，第二个位置有 4 种选择，依此类推。

n! = n × (n − 1) × (n − 2) × … × 1

5! = 5 × 4 × 3 × 2 × 1 = 120

3. Choosing Ministers Without Order: Combinations | 无序选择部长：组合

A combination is a selection of items where the order does not matter. The number of ways to choose r items from n distinct items is written as ⁿCᵣ or C(n, r).

组合是指不考虑顺序的选择。从 n 个不同物品中选出 r 个的方式数记为 ⁿCᵣ 或 C(n, r)。

ⁿCᵣ = n! ÷ [r! (n − r)!]

ⁿCᵣ = n! / [r! (n − r)!]

Example: From a party of 10 MPs, a cabinet of 3 ministers is to be selected without assigning specific portfolios. The number of possible cabinets is ¹⁰C₃ = 10! / (3! × 7!) = 120.

例题：从 10 名议员中选出 3 名部长，但不分配具体职位。可能的内阁数为 ¹⁰C₃ = 10! / (3! × 7!) = 120。

English: Order does not matter in a combination.	中文：组合中顺序不重要。
English: ⁿCᵣ = ⁿCₙ₋ᵣ, so ¹⁰C₃ = ¹⁰C₇.	中文：ⁿCᵣ = ⁿCₙ₋ᵣ，因此 ¹⁰C₃ = ¹⁰C₇。

4. Assigning Portfolios: Permutations | 分配部长职位：排列

A permutation is an arrangement of items where the order matters. The number of ways to arrange r items chosen from n distinct items is written as ⁿPᵣ.

排列是指考虑顺序的排列方式。从 n 个不同物品中选出 r 个并排序的方式数记为 ⁿPᵣ。

ⁿPᵣ = n! ÷ (n − r)!

ⁿPᵣ = n! / (n − r)!

If a cabinet has 3 distinct portfolios such as Finance, Defence and Health, and there are 10 available MPs, then the number of ways to fill these posts is ¹⁰P₃ = 10 × 9 × 8 = 720.

如果内阁有 3 个不同职位，例如财政大臣、国防大臣和卫生大臣，且有 10 名议员可供选择，那么填补这些职位的排列数为 ¹⁰P₃ = 10 × 9 × 8 = 720。

This is larger than the corresponding combination because each group of 3 selected MPs can be assigned to the 3 portfolios in 3! = 6 different ways.

这比对应的组合数大，因为每组被选中的 3 名议员可以以 3! = 6 种不同方式分配到 3 个职位上。

5. The Cabinet as a Committee: Combinations in Politics | 内阁作为委员会：政治中的组合

Many Edexcel exam questions model committees, teams or cabinets as unordered selections. Once you see the phrase ‘select a group’ or ‘choose a committee’, you should think of combinations.

许多 Edexcel 考试题将委员会、团队或内阁建模为无序选择。一旦看到 “选择一个小组” 或 “选出委员会” 这样的表述，就应该想到组合。

Suppose a Prime Minister must choose 5 cabinet members from a pool of 12 loyal MPs. The number of possible 5-member cabinets is ¹²C₅.

假设首相必须从 12 名忠诚议员中选出 5 名内阁成员。可能的 5 人内阁数量为 ¹²C₅。

¹²C₅ = 12! / (5! × 7!) = 792

If the 5 cabinet posts are all equal in status, then order is irrelevant. Once portfolios such as Chancellor and Home Secretary are assigned, the problem becomes a permutation.

如果这 5 个内阁职位地位相同，那么顺序无关紧要。一旦分配财政大臣、内政大臣等具体职位，问题就变成了排列。

6. Total Cabinets from a Party List | 从政党名单组阁

When forming a cabinet from a larger party list, the total number of possible cabinets depends on whether all members are distinct and whether posts are distinct.

从较大的政党名单中组阁时，可能的内阁总数取决于成员是否互不相同以及职位是否互不相同。

Case 1: Select 4 equal ministers from 15 MPs. Number of ways = ¹⁵C₄.

情况 1：从 15 名议员中选出 4 名地位相同的部长。方式数 = ¹⁵C₄。

Case 2: Assign 4 distinct portfolios from the same 15 MPs. Number of ways = ¹⁵P₄ = 15 × 14 × 13 × 12.

情况 2：从同样的 15 名议员中分配 4 个不同职位。方式数 = ¹⁵P₄ = 15 × 14 × 13 × 12。

Scenario 场景	Order matters? 顺序是否重要？	Formula 公式
Equal cabinet posts 平等内阁职位	No 否	ⁿCᵣ
Distinct portfolios 不同职位	Yes 是	ⁿPᵣ

7. Restricted Selections: Must Include the Prime Minister | 受限选择：必须包括首相

Restricted selection problems require you to fix certain members first, then count the remaining choices separately.

受限选择问题要求先固定某些成员，然后单独计算剩余的选择。

Example: A cabinet of 4 is to be selected from 9 MPs. The current Prime Minister must be included. Since the PM is already selected, choose the remaining 3 members from the other 8 MPs.

例题：从 9 名议员中选出 4 人组成内阁，现任首相必须包括在内。由于首相已入选，只需从其余 8 名议员中选出 3 人。

Number of ways = ⁸C₃ = 8! / (3! × 5!) = 56

方式数 = ⁸C₃ = 8! / (3! × 5!) = 56

If the Prime Minister must not be included, then select all 4 members from the remaining 8 MPs, giving ⁸C₄ = 70 ways.

如果首相不得入选，则从其余 8 名议员中选出全部 4 人，方式数为 ⁸C₄ = 70。

8. The Multinomial Coefficient: Splitting Portfolios | 多项式系数：分配多个职位类别

When dividing a group into more than two categories, the multinomial coefficient generalises the binomial coefficient. If n distinct people are split into groups of sizes k₁, k₂, k₃, …, the number of ways is:

当把一个群体分成两个以上类别时，多项式系数推广了二项式系数。如果 n 个不同的人被分成大小为 k₁、k₂、k₃、… 的组，方式数为：

n! / (k₁! × k₂! × k₃! × …)

Example: From 15 MPs, appoint 3 to Finance, 4 to Defence and 8 to Backbench roles. The number of distinct allocations is 15! / (3! × 4! × 8!).

例题：从 15 名议员中，任命 3 人到财政委员会、4 人到国防委员会、8 人担任后座议员。不同分配方式数为 15! / (3! × 4! × 8!)。

This is because the three groups are labelled by their roles. If the groups were unlabelled, you would divide further by the number of identical group sizes.

这是因为三个组是按职责标记的。如果组没有标记，则还需要除以相同大小组的排列数。

9. Probability of a Particular Cabinet | 特定内阁的概率

If all possible cabinets are equally likely, the probability of a particular cabinet being selected is 1 divided by the total number of possible cabinets.

如果所有可能的内阁等可能，则某一特定内阁被选中的概率为 1 除以可能的内阁总数。

Example: A 3-member cabinet is selected at random from 10 MPs. What is the probability that MPs A, B and C are chosen?

例题：从 10 名议员中随机选出 3 人内阁。议员 A、B、C 被选中的概率是多少？

Total possible cabinets = ¹⁰C₃ = 120

可能的内阁总数 = ¹⁰C₃ = 120

P(A, B, C selected) = 1 / 120

P(A、B、C 被选中) = 1 / 120

If the question asks for the probability that A is selected but B is not, count the favourable outcomes using restricted combinations, then divide by the total.

如果问题问 A 被选中但 B 未被选中的概率，则用受限组合计算有利结果数，再除以总数。

10. Realistic Constraints and Overcounting | 现实约束与重复计数

Real cabinet formation often includes constraints such as ‘at least one woman’ or ‘no more than two from the same faction’. These are handled by splitting into cases and adding mutually exclusive outcomes.

现实组阁常包含约束，如 “至少一名女性” 或 “同一派系不超过两人”。处理方法是分情况讨论，并相加互斥的结果。

For example, a 4-person cabinet must include at least one female MP. If there are 6 male and 5 female MPs, count the cases of exactly 1, 2, 3 or 4 women.

例如，一个 4 人内阁必须至少包含一名女性议员。若有 6 名男性和 5 名女性议员，则分别计算恰好 1、2、3、4 名女性的情况。

Alternatively, use the complement: total ways without restriction minus ways with zero women.

或者使用补集：无限制的总方式数减去没有女性的方式数。

Total = ¹¹C₄; Zero women = ⁶C₄

总数 = ¹¹C₄；没有女性 = ⁶C₄

At least one woman = ¹¹C₄ − ⁶C₄ = 330 − 15 = 315

至少一名女性 = ¹¹C₄ − ⁶C₄ = 330 − 15 = 315

11. Exam-Style Question Walkthrough | 考试题型解析

Question: A Prime Minister is to choose a cabinet of 6 ministers from 14 available MPs. Among the 14 MPs, 8 are from the senior wing and 6 are from the junior wing. The cabinet must contain exactly 4 senior MPs and 2 junior MPs.

问题：首相要从 14 名议员中选出 6 人内阁，其中 8 人来自资深翼，6 人来自新晋翼。内阁必须恰好包含 4 名资深议员和 2 名新晋议员。

Step 1: Choose the 4 senior MPs from 8. This can be done in ⁸C₄ ways.

步骤 1：从 8 名资深议员中选出 4 人，有 ⁸C₄ 种方式。

Step 2: Choose the 2 junior MPs from 6. This can be done in ⁶C₂ ways.

步骤 2：从 6 名新晋议员中选出 2 人，有 ⁶C₂ 种方式。

⁸C₄ = 70 and ⁶C₂ = 15

⁸C₄ = 70，⁶C₂ = 15

Step 3: Multiply the independent choices. Total ways = 70 × 15 = 1050.

步骤 3：将相互独立的选择相乘。总方式数 = 70 × 15 = 1050。

This two-stage multiplication is a common Edexcel exam technique. Always check that the categories do not overlap and that every cabinet is counted exactly once.

这种两阶段乘法是 Edexcel 考试的常用技巧。务必检查各分类不重叠，且每个内阁恰好被计数一次。

12. Common Mistakes and Key Takeaways | 常见错误与关键要点

One common mistake is using permutations when order does not matter, which leads to overcounting by a factor of r!.

一个常见错误是在顺序无关时使用了排列，导致重复计数 r! 倍。

Another error is treating restricted selections as a single combination without fixing the required members first. Always deal with mandatory or forbidden members before counting the rest.

另一个错误是将受限选择当作单一组合处理，而没有先固定必须包括或排除的成员。务必先处理强制或禁止成员，再计算其余部分。

English: Use ⁿCᵣ for ‘select’ or ‘choose’ when order does not matter.
中文：当顺序无关，遇到 “选择” 或 “选出” 时使用 ⁿCᵣ。
English: Use ⁿPᵣ for ‘arrange’ or ‘assign distinct roles’.
中文：当涉及 “排列” 或 “分配不同职位” 时使用 ⁿPᵣ。
English: For restricted problems, split into cases or use the complement.
中文：对于受限问题，分情况讨论或使用补集。
English: Always check whether groups are labelled before applying the multinomial coefficient.
中文：在应用多项式系数前，务必检查各组是否有标记。

Mastering these counting principles will help you tackle Edexcel Pure Mathematics questions involving selections, arrangements and probability with confidence.

掌握这些计数原理将帮助你自信应对 Edexcel 纯数学中涉及选择、排列和概率的问题。

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The Cabinet: Permutations, Combinations and Ministerial Selections | 内阁：排列组合与部长选拔