📚 Exercise 21A.2: Counting Principle Problem | 练习21A.2：计数原理问题
Combinatorics and counting principles form a foundational pillar of the IB Mathematics syllabus, appearing in both Standard Level and Higher Level papers. Exercise 21A.2 typically challenges students to apply permutations or combinations under a restriction—often a “not together” condition. Mastering these problems builds logical reasoning and reinforces the importance of systematic counting. This article breaks down a classic committee-selection problem with a refusal constraint, demonstrating multiple solution methods, common pitfalls, and links to IB assessment objectives.
组合数学和计数原理是IB数学课程的基础支柱，既出现在标准级别也出现在高级别考卷中。练习21A.2通常让学生在一个限制条件下应用排列或组合——常见的是“不能同时出现”的条件。掌握这类题目能够培养逻辑推理能力，并强化系统计数的重要性。本文剖析一个经典的、带拒绝条件的委员会选题，展示多种解法、常见错误以及与IB评估目标的联系。
1. Problem Statement | 问题陈述
A committee of 5 members is to be selected from a group of 12 people. Two specific individuals, Alice and Bob, refuse to work together. How many different committees can be formed? This is a typical “not both” restriction applied to combinations.
从12人中选出一个5人委员会。其中特定的两人Alice和Bob拒绝一起工作。问有多少种不同的委员会组成方式？这是一个应用于组合的典型“不能两人同时”限制。
2. Understanding the Key Condition | 理解关键条件
The phrase “refuse to work together” means that a committee containing both Alice and Bob is not allowed. Committees that include exactly one of them, or neither, are perfectly valid. This transforms a straightforward combination C(12,5) into a conditional counting exercise. We must ensure that every committee we count does not violate the restriction.
“拒绝一起工作”意味着包含Alice和Bob两人的委员会是不被允许的。只包含其中一人的委员会，或两人都不包含，是完全有效的。这将一个直接的组合C(12,5)变成了一个条件计数练习。我们必须确保计数的每个委员会都不违反限制。
3. Counting Principles Review | 计数原理回顾
distinct items when order does not matter is given by the combination formula: C(n, r) = n! / [r! (n - r)!]. For example, C(12,5) = 12! / (5! × 7! rmutation (order matters), we would use P(n, r), but here a committee is an unordered set.
在IB数学中，从n个不同物品中无序选取r个的方法数由组合公式给出：C(n, r) = n! / [r! (n - r)!]。例如，C(12,5) = 12! / (5! × 7!) = 792，即总的无限制委员会数。如果问题涉及排列（顺序重要），我们会使用P(n, r)，但这里的委员会是无序集合。
C(12,5) = 12! / (5! × 7!) = 792
C(12,5) = 12! / (5! × 7!) = 792
4. Method 1: Complementary Counting | 方法一：互补计数
Complementary counting is often the most efficient technique for “not both” conditions. We first count the total number of unrestricted committees, then subtract those that violate the rule. A violating committee must contain both Alice and Bob. If both are already chosen, we need 3 more members from the remaining 10 people (excluding Alice and Bob). Thus, the number of forbidden committees is C(10,3).
互补计数通常是处理“不能两人同时”条件最有效的方法。我们先计算无限制的总委员会数，然后减去违反规则的委员会。违反规则的委员会必定同时包含Alice和Bob。如果两人已被选中，我们需要从剩下的10人（排除Alice和Bob）中再选3人。因此，禁止委员会的数量为C(10,3)。
The allowed committees are then: Total – Forbidden = 792 – C(10,3).
被允许的委员会数为：总数 - 禁止数 = 792 – C(10,3)。
It is crucial to recognize that the “forbidden” set must be subtracted only once. Because the condition is simply “both cannot be together”, there is no overlap between forbidden and allowed sets, making subtraction straightforward.
关键是要认识到“禁止”集合只需减去一次。因为条件只是“两人不能同时出现”，禁止集与允许集之间没有重叠，减法直接有效。
5. Step-by-Step Calculation (Method 1) | 逐步计算（方法一）
Calculate C(10,3) = 10! / (3! × 7! is gives 120 committees that contain both Alice and Bob. Therefore, valid committees = 792 – = 672. You should always check mental arithmetic, especially in non-calculator IB papers.
计算 C(10,3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120。即包含Alice和Bob两人的委员会有120个。因此，有效委员会 = 792 – 120 = 672。尤其在IB非计算器试卷中，要始终检查心算。
C(10,3) = 120, Valid committees = 672
C(10,3) = 120, 有效委员会数 = 672
6. Method 2: Direct Case Analysis | 方法二：直接分类分析
For a deeper understanding—and to solidify the logic—you can break the selection into three mutually exclusive cases based on the presence of Alice and Bob. Since these cases cover all possibilities without overlap, the total number of valid committees is their sum. The cases are:
为了更深入地理解，并巩固逻辑，你可以根据Alice和Bob的出现情况将选择分解为三个互斥的情形。因为这些情形涵盖了所有可能性且无重叠，有效委员会总数就是它们的和。情形如下：
e and exclude Bob; choose 4 others from the remaining 10 people. Number of ways = C(10,4).
情形(i)：Alice被选中，Bob不被选中。固定Alice并排除Bob；从其余10人中选4人。方法数 = C(10,4)。
Case (ii): Bob is selected, but Alice is not. Symmetrically, also C(10,4).
情形(ii)：Bob被选中，Alice不被选中。对称地，也是C(10,4)。
elected. Choose all 5 members from the 10 people excluding both. Number of ways = C(10,5).
情形(iii)：Alice和Bob都不被选中。从排除两人的10人中选出全部5人。方法数 = C(10,5)。
7. Step-by-Step Calculation (Method 2) | 逐步计算（方法二）
Now compute the values: C(10,4) = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 210. C(10,5) = 10! / (5! × 5! × 2 × 1) = 252. Adding them: 210 + 210 + 252 = 672. The result matches Method 1 perfectly.
计算这些值：C(10,4) = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 210。C(10,5) = 10! / (5! × 5!) = (10 × 9 × 8 × 7 × 6) / (5 × 4 × 3 × 2 × 1) = 252。相加得：210 + 210 + 252 = 672。结果与方法一完全吻合。
This casework approach not only confirms the answer but also demonstrates systematic problem-solving, which is highly valued in IB exams. Should the restriction become more complex—e.g., three people with pairwise conflicts—extension of this casework becomes essential.
这种分类讨论方法不仅证实了答案，还展示了系统的解决问题能力，这在IB考试中备受重视。如果限制变得更复杂，例如三人之间的两两冲突，推广这种分类讨论就变得至关重要。
8. Verification and Consistency Check | 验证与一致性检查
cases gives strong confidence. Additionally, you can verify with a general identity: For a “not both A and B” restriction, the number of valid r-combinations from n people is C(n,r) – C(n-2, r-2), provided n ≥ 2 and r ≥ 2. Substituting n=12, r=5: C(12,5) – C(10,3) = 792 – ng 2 from 4 where two people refuse to be together, to ensure your logic scales correctly.
互补计数
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