The Work of Select Committees: Combinatorics in Action | 选择委员会的工作：组合数学应用

📚 The Work of Select Committees: Combinatorics in Action | 选择委员会的工作：组合数学应用

In Edexcel A-Level Mathematics, many probability and statistics questions are built around the idea of forming a select committee from a larger group. Although the real-world work of select committees belongs to politics, the mathematical modelling of how such committees are chosen is a rich source of exam problems. This article breaks down the combinatorial and probabilistic techniques needed to solve committee selection questions, from basic counting to the hypergeometric distribution.

在 Edexcel A-Level 数学中，许多概率与统计题目都以从较大群体中组建一个选择委员会为背景。尽管现实中的 select committee 工作属于政治领域，但如何用数学模型描述此类委员会的产生方式，是考试中常见的题型来源。本文梳理解决委员会选择问题所需的组合与概率方法，从基础计数到超几何分布。

1. Counting Principles for Committee Selection | 委员会选择中的计数原理

When forming a select committee, we usually ask: in how many ways can we choose the members? The multiplicative principle states that if one task can be done in m ways and a second independent task in n ways, then both can be done in m × n ways. The additive principle applies when we count disjoint alternatives, so we add the numbers of ways for each case.

组建一个选择委员会时，我们通常要问：有多少种方式可以选出成员？乘法原理指出，若一项任务有 m 种完成方式，另一项独立任务有 n 种方式，则两项任务共同完成有 m × n 种方式。加法原理适用于互斥的不同情形，此时我们把每种情形的数量相加。

For successive choices, multiply the number of options at each step.
For alternative cases that cannot overlap, add the separate counts.
Always check whether choices are independent or mutually exclusive before applying the rule.

对于连续的选择，每一步选项数相乘。
对于互不重叠的备选情形，将各情形的计数相加。
在应用原理前，先判断选择之间是相互独立还是互斥。

2. Permutations vs Combinations | 排列与组合

A permutation is an arrangement where order matters, while a combination is a selection where order does not matter. For a committee, the order in which members are chosen is irrelevant, so combinations are used. However, if the committee assigns distinct roles such as chair, secretary and treasurer, then order or assignment matters, and permutations apply.

排列是一种顺序重要的排列方式，而组合是一种顺序无关紧要的选择方式。对于委员会成员，选择顺序并不重要，因此使用组合。但如果委员会要分配主席、秘书、财务等不同职务，那么顺序或任务分配就重要，此时使用排列。

Scenario	Order matters?	Method
Selecting 4 members from 10 people	No	Combination
Choosing a chair, secretary and treasurer from 10 people	Yes	Permutation

Permutations: P(n, r) = n! ÷ (n − r)!

Combinations: C(n, r) = n! ÷ (r!(n − r)!)

3. The Combination Formula | 组合公式

The number of ways to choose r objects from n distinct objects without regard to order is written as C(n, r), also denoted ⁿCᵣ or binom(n,r). The formula is built from the permutation count divided by r! to remove the effect of ordering the chosen items.

从 n 个不同对象中不考虑顺序地选出 r 个对象的方法数记作 C(n, r)，也写作 ⁿCᵣ 或 binom(n,r)。该公式由排列数除以 r! 得到，以消除所选对象的顺序影响。

C(n, r) = n! ÷ (r!(n − r)!)

For example, C(10, 4) = 10! ÷ (4! × 6!) = (10 × 9 × 8 × 7) ÷ (4 × 3 × 2 × 1) = 210. This tells us that from 10 people, we can form 210 different committees of size 4.

例如，C(10, 4) = 10! ÷ (4! × 6!) = (10 × 9 × 8 × 7) ÷ (4 × 3 × 2 × 1) = 210。这意味着从 10 人中可以组成 210 个不同的 4 人委员会。

C(n, 0) = 1 and C(n, n) = 1.
C(n, r) = C(n, n − r) — choosing r is the same as leaving out n − r.
Use the factorial formula when numbers are small; for large n, use a calculator or Pascal’s triangle.

C(n, 0) = 1，C(n, n) = 1。
C(n, r) = C(n, n − r) —— 选出 r 个等价于留下 n − r 个。
数字较小时用阶乘公式；n 较大时使用计算器或帕斯卡三角形。

4. Selecting a Basic Committee | 选择基本委员会

The simplest committee question gives a group size n and a committee size r, and asks for the number of possible committees. Since order does not matter, the answer is simply C(n, r). This forms the denominator in many probability questions.

最基本的委员会题目给出总人数 n 和委员会人数 r，要求计算可能的委员会数量。由于顺序无关，答案就是 C(n, r)。在许多概率题中，这个数构成分母。

Worked example: A club has 12 members. How many ways can a committee of 5 be chosen?

例题：某俱乐部有 12 名成员，有多少种方式选出 5 人委员会？

C(12, 5) = 12! ÷ (5! × 7!) = 792

Therefore, there are 792 possible committees. If the question asks for the probability that a particular member is selected, we can count favourable outcomes by fixing that member and choosing the remaining 4 from the other 11, giving C(11, 4) = 330. The probability is 330 ÷ 792 = 5/12.

因此共有 792 个可能的委员会。若题目问某位特定成员被选中的概率，我们可以固定该成员，然后从其余 11 人中选出剩余 4 人，得到 C(11, 4) = 330。概率为 330 ÷ 792 = 5/12。

5. Committees with Restrictions | 带限制条件的委员会

Many exam questions add conditions such as ‘at least one woman’, ‘exactly two men’, or ‘the chair must be a woman’. The key is to split the group into sub-groups and apply the multiplication principle within each case, then use the addition principle across disjoint cases.

许多考题会附加条件，例如“至少一名女性”“恰好两名男性”或“主席必须是女性”。关键是把总体分成若干子组，在每种情形内部应用乘法原理，然后在互斥情形之间应用加法原理。

Example: From 8 men and 6 women, choose a committee of 5 with exactly 2 men.

例题：从 8 名男性和 6 名女性中选出 5 人委员会，要求恰好有 2 名男性。

Number of ways = C(8, 2) × C(6, 3) = 28 × 20 = 560

For ‘at least one woman’, it is often faster to subtract the all-male committee from the total number of committees.

对于“至少一名女性”，通常用总数减去全为男性的委员会数量更快。

Total committees C(14, 5) = 2002; all-male C(8, 5) = 56; so at least one woman = 2002 − 56 = 1946

Identify disjoint cases first, then add their counts.
When ‘at least’ appears, consider the complement: 1 − P(none).
Double-check that each case is mutually exclusive to avoid double counting.

先找出互斥情形，再将其计数相加。
出现“至少”时，可考虑补集：1 − P(无此类)。
检查各情形是否互斥，避免重复计数。

6. Probability of Selecting a Committee | 选择委员会的概率

When all committees of a fixed size are equally likely, the probability of a particular type of committee is the number of favourable committees divided by the total number of committees. This is the classical probability formula.

当所有固定大小的委员会等可能时，某类委员会的概率等于有利委员会的数量除以委员会总数。这就是古典概率公式。

P(event) = Number of favourable committees ÷ Total number of committees

Example: A committee of 4 is chosen at random from 7 men and 5 women. Find the probability that the committee has exactly 3 women.

例题：从 7 名男性和 5 名女性中随机选出 4 人委员会，求恰好有 3 名女性的概率。

Total = C(12, 4) = 495; favourable = C(5, 3) × C(7, 1) = 10 × 7 = 70

P(exactly 3 women) = 70 ÷ 495 = 14/99

Always state the total number of equally likely outcomes first. If the selection is not equally likely, use a probability tree or distribution instead.

先说明等可能结果的总数。如果选择不是等可能的，则改用概率树或分布。

7. Hypergeometric Distribution | 超几何分布

Committee selection without replacement from two categories (e.g. men and women) leads directly to the hypergeometric distribution. If a committee of size n is drawn from a group of a items of type A and b items of type B, then the probability of exactly k items of type A is:

从两类对象（如男性和女性）中不放回地抽取委员会，直接对应超几何分布。若从 a 个 A 类对象和 b 个 B 类对象中抽取规模为 n 的委员会，则恰好有 k 个 A 类对象的概率为：

P(X = k) = C(a, k) × C(b, n − k) ÷ C(a + b, n)

Here X is the number of type A members in the committee. The denominator C(a + b, n) is the total number of possible committees, while the numerator counts committees with exactly k type A and n − k type B members.

这里 X 是委员会中 A 类成员的数量。分母 C(a + b, n) 是可能委员会的总数，分子则计数恰好有 k 个 A 类成员和 n − k 个 B 类成员的委员会。

The hypergeometric distribution models sampling without replacement.
Use it when the population is small and finite, and items are drawn without replacement.
The support of X is max(0, n − b) ≤ k ≤ min(a, n).

超几何分布描述不放回抽样。
当总体有限且较小，并且不放回抽取时使用。
X 的取值范围是 max(0, n − b) ≤ k ≤ min(a, n)。

8. Expected Value and Variance | 期望与方差

For a hypergeometric random variable X representing the number of type A members in a committee of size n drawn from a total of N = a + b items, the expected value and variance have simple forms.

对于超几何随机变量 X，表示从总共 N = a + b 个对象中抽取的规模为 n 的委员会中 A 类成员的数量，其期望和方差具有简洁形式。

E(X) = n × (a ÷ (a + b))

Var(X) = n × (a ÷ (a + b)) × (b ÷ (a + b)) × ((a + b − n) ÷ (a + b − 1))

Notice that E(X) is the same as the binomial expectation for a proportion a/(a+b), but the variance is smaller because the population is finite and sampling is without replacement. This reduction is captured by the finite population correction factor (N − n)/(N − 1).

注意 E(X) 与二项分布中比例 a/(a+b) 的期望相同，但方差更小，因为总体有限且不放回抽样。这一减小由有限总体修正因子 (N − n)/(N − 1) 体现。

9. Binomial Approximation | 二项近似

When the total population is very large compared with the committee size, the hypergeometric distribution can be approximated by a binomial distribution with parameters n and p = a/(a+b). This is useful because binomial calculations are often simpler, but for small groups the hypergeometric model is exact.

当总人口数远大于委员会规模时，超几何分布可以用参数为 n 和 p = a/(a+b) 的二项分布近似。这很有用，因为二项计算通常更简单，但对于小群体，超几何模型才是精确的。

If N is large and n/N < 0.1, then X ~ B(n, p) approximately, where p = a/(a+b)

However, for typical A-Level committee questions, the population sizes are small enough that you should use the hypergeometric formula directly. Always check whether replacement occurs: with replacement use binomial, without replacement use hypergeometric.

然而，在典型的 A-Level 委员会题目中，总体规模足够小，应直接使用超几何公式。始终确认是否有放回：有放回用二项分布，无放回用超几何分布。

With replacement → binomial distribution.
Without replacement and finite population → hypergeometric distribution.
If n/N is very small, the hypergeometric probabilities are close to binomial probabilities.

有放回 → 二项分布。
无放回且总体有限 → 超几何分布。
若 n/N 非常小，超几何概率接近二项概率。

10. Exam Tips and Common Mistakes | 考试技巧与常见错误

Committee selection problems appear regularly in Edexcel A-Level Mathematics, especially in Statistics units. Follow these tips to maximise marks:

委员会选择问题在 Edexcel A-Level 数学的统计单元中经常出现。遵循以下建议以获得最高分数：

Read the question carefully: identify the total group size, committee size, and any restrictions.
Decide whether order matters. Committees usually use combinations, not permutations.
Write down the total number of equally likely committees before computing probabilities.
For ‘at least one’ conditions, use the complement when it is faster.
Check your arithmetic: factorials and C(n, r) values can be large, so use the calculator correctly.
Label all random variables and state the distribution clearly, e.g. X ~ Hypergeometric(N, a, n).

仔细读题：确定总人数、委员会规模以及任何限制条件。
判断顺序是否重要。委员会通常使用组合而非排列。
在计算概率前，先写出等可能委员会的总数。
对于“至少一个”条件，若补集更快则使用补集。
检查计算：阶乘和 C(n, r) 值可能很大，正确使用计算器。
标注所有随机变量并清楚写出分布，例如 X ~ Hypergeometric(N, a, n)。

Common mistakes include double counting overlapping cases, using permutations instead of combinations, forgetting the complement, and applying the binomial distribution when sampling is without replacement from a small finite group.

常见错误包括重复计数重叠情形、误用排列代替组合、忘记补集，以及在小有限群体不放回抽样时错误应用二项分布。

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The Work of Select Committees: Combinatorics in Action | 选择委员会的工作：组合数学应用