📚 The Cabinet (Pigeonhole Principle) | 鸽巢原理
The 'cabinet' principle is a deceptively simple idea with powerful consequences across the Edexcel A-Level Mathematics syllabus. Often called the pigeonhole principle, it states that if you try to place more items than there are cabinets, at least one cabinet must contain at least two items. This idea emerges in statistics, decision mathematics, and even pure problem-solving contexts, providing a logical shortcut to prove that certain patterns are unavoidable.
“柜子”原理（也称鸽巢原理）是一个看似简单却在Edexcel A-Level数学大纲中有着强大威力的思想。它指出，如果你想把比柜子数量更多的物品放进柜子，那么至少有一个柜子必须容纳至少两件物品。这个想法出现在统计、决策数学甚至纯解题情境中，为证明某些模式不可避免提供了一条逻辑捷径。
1. The Simple Form of the Principle | 原理的简单形式
If n items are placed into m cabinets and n > m, then there exists at least one cabinet that contains at least two items. This is the simplest statement of the cabinet principle, and it is surprisingly useful. For example, among any 13 people, since there are only 12 months in a year, the principle guarantees that at least two people share the same birth month. No data collection is needed – the conclusion follows from pure logic.
如果将 n 件物品放入 m 个柜子，且 n > m，则至少存在一个柜子至少包含两件物品。这是柜子原理最简单的陈述，但用处惊人。例如，任意 13 个人中，因为一年只有 12 个月，该原理就能保证至少有两个人的生日月份相同。无需收集数据——结论直接由纯逻辑得出。
2. Representing the Principle Mathematically | 用数学语言表达原理
We can formalise the cabinet principle using set notation. Let the set of items be I, with |I| = n, and the set of cabinets be C, with |C| I → C. If n > m, the function cannot be injective; i.e., there must exist distinct i1, i2 ∈ such that f(i1) = f(i2). In plain words, two different items must go to the same cabinet.
我们可以用集合论的语言形式化柜子原理。设物品集合为 I，|I| = n，柜子集合为 C，|C| = m。放置方式即为函数 f: I → C。若 n > m，该函数不可能为单射，即必定存在不同的 i1, i2 ∈ I 使得 f(i1) = f(i2)。通俗地说，两个不同的物品必定进入同一个柜子。
3. Proof by Contradiction | 反证法证明
A standard proof of the cabinet principle uses contradiction. Assume the opposite: that each cabinet holds at most one item. Then the total number of items placed would be at most m. But we are given n > m, so this is impossible. Hence, the assumption fails, and at least one cabinet must hold two or more items. This short proof is often required in Edexcel exam questions to demonstrate logical rigour.
柜子原理的标准证明采用反证法。假设每个柜子最多放一件物品，那么总共最多只能放置 m 件物品。但已知 n > m，因此不可能。所以原假设不成立，至少有一个柜子必须容纳两件或更多物品。这个简短的证明在Edexcel考试题中经常被要求，以展示逻辑严密性。
4. The Generalized Pigeonhole Principle | 广义鸽巢原理
es: if n items are placed into m cabinets, then there exists a cabinet containing at least ⌈n/m⌉ items, where ⌈ ⌉ otes the ceiling function. For instance, if 100 students take 4 optional modules, at least ⌈100/4⌉ e. This generalised form is extremely handy when we need a minimum guaranteed number rather than just 'at least two'.
更强的版本指出：若将 n 件物品放入 m 个柜子，则存在某个柜子至少包含 ⌈n/m⌉ 件物品，其中 ⌈ ⌉ 表示上取整函数。例如，100 名学生选择 4 个选修模块，则至少有 ⌈100/4⌉ = 25 名学生必定选择了同一个模块。当我们需要保证的最小数目而不仅仅是“至少两个”时，这个广义形式非常有用。
5. Application in Number Theory | 数论中的应用
The cabinet principle shines in number theory problems featured in Further Mathematics. A classic example: given any 5 integers, prove that at least two have a difference divisible by 4. Here, cabinets are the possible remainders when divided by 4: {0, 1, 2, 3} — that's 4 cabinets. The 5 integers are the items. Since 5 > 4, two numbers must share the same remainder, so their difference is a multiple of 4.
柜子原理在进阶数学的数论问题中大放异彩。一个经典例子：任给 5 个整数，证明至少有两个数之差可被 4 整除。这里，柜子是被 4 除时可能的余数集合：{0, 1, 2, 3}，共 4 个柜子；物品就是这 5 个整数。由于 5 > 4，必有两个数具有相同的余数，因此它们的差是 4 的倍数。
6. Combinatorial Selection Problems | 组合选择问题
Edexcel statistics and decision papers often ask questions about unavoidable combinations. Consider choosing 6 numbers from the set {1, 2, ..., 10}. Prove that at least two chosen numbers must sum to 11. The cabinet pairs that sum to 11 are (1,10), (2,9), (3,8), (4,7), (5,6). These 5 pairs form 5 cabinets. Choosing 6 numbers forces at least one pair to be fully selected, giving the required sum.
Edexcel 的统计和决策试卷常会问及不可避免的组合问题。考虑从集合 {1, 2, ..., 10} 中选择 6 个数字，证明至少有两个被选数字之和为 11。和为 11 的的“抽屉对”为 (1,10), (2,9), (3,8), (4,7), (5,6)，这 5 对构成 5 个柜子。选出 6 个数字必然会导致至少有一对被完整选中，从而得到所需的和。
7. Geometric Applications | 几何中的应用
nit squares. If 5 points are placed inside the big square, at least two points must lie inside the same small square, so the distance between them is at most the diagonal of the small square, which is √2. This elegant argument uses area as cabinets and shows existence without needing coordinates.
在几何情境中，该原理可以证明邻近性。考虑边长为 2 的正方形，划分为四个 1 × 1 的单位正方形。如果在大正方形内部放入 5 个点，则至少有两个点必定落在同一个小正方形内，因此它们之间的距离至多为小正方形的对角线长 √2。这个优雅的论证将面积当作柜子，无需坐标就能证明存在性。
8. Probability and the Cabinet Principle | 概率与柜子原理
While the cabinet principle guarantees a 'sure' event, in probability we can use it to determine when an event becomes certain. For instance, in a bag of 8 red and 6 blue sweets, the smallest number you must pick to be certain of getting at least two of the same colour is 3 (by the pigeonhole principle with two colours). Edexcel exam questions may combine this with probability calculations to test deeper understanding.
虽然柜子原理保证的是“必然”事件，但在概率中我们可以用它来判断何时某事件变得确定。例如，袋中有 8 颗红色和 6 颗蓝色糖果，为保证至少拿到两颗同色，你必须最少取出 3 颗（由鸽巢原理以两种颜色得出）。Edexcel 的考题可能会将此与概率计算结合，以检验更深层次的理解。
9. Typical Edexcel Exam-Style Questions | 典型 Edexcel 考题风格
A common question asks: 'e that in any group of 8 people, at least two have a birthday on the same day of the week.' ize to guarantee at least 3 people share a weekday birthday: using the generalised form, with 7 days, we need n such that ⌈n/7⌉ = 2x7 + 1 = 15. Such structured problem-solving appears in Statistics and Decision papers.
一道常见的题目是：“证明在任何 8 人的群体中，至少有两人在一周的同一天过生日。”解法将 7 天作为柜子，8 人作为物品，直接应用简单形式。更进阶的问题可能会问：为保证至少有 3 人在同一工作日出生的最小群体规模。使用广义形式，以 7 天计算，需要 n 使得 ⌈n/7⌉ ≥ 3，得 n = 2×7 + 1 = 15。这类结构化的问题解决出现在统计与决策试卷中。
10. Avoiding Common Pitfalls | 避免常见陷阱
Students often misidentify cabinets or items. Always ask: what is being assigned to what? The smaller number is usually the number of cabinets. Another mistake is forgetting the ceiling function in the generalised form and rounding down instead. A third error is using the principle where it does not apply because the assignment is not arbitrary. In Edexcel mark schemes, clear identification of cabinets and items is essential for marks.
学生常常会错误地分辨柜子和物品。始终要问：将什么分配给什么？较小的数字通常是柜子的数量。另一个错误是在广义形式中忘记上取整函数，错误地向下取整。第三个错误是在分配并非任意时将原理用于不适用的情况。在 Edexcel 的评分方案中，清晰地指出柜子和物品对于得分至关重要。
11. The Principle in Decision Mathematics | 决策数学中的柜子原理
In Edexcel's Decision Mathematics modules, the cabinet principle underpins algorithms and network proofs. For example, in scheduling problems with more tasks than processors, at least one processor must handle multiple tasks. Similarly, the proof that a simple graph with n vertices and no multiple edges must have at least two vertices with the same degree uses the cabinet principle with n-1 possible degree values (0 to n-1), but degree 0 and degree n-1 cannot coexist, reducing cabinets to n-1 for n items, guaranteeing a repeat.
在 Edexcel 的决策数学模块中，柜子原理支撑着算法与网络的证明。例如，在任务数多于处理器的调度问题中，至少有一个处理器必须处理多项任务。同样，证明无重边的 n 阶简单图至少有两个顶点具有相同度数的过程，使用了柜子原理：可能的度数为 0 到 n-1，共 n 种，但度 0 和度 n-1 不能共存，因此实际只有 n-1 个柜子，对 n 个物品来说必然有重复。
12. Summary and Key Takeaways | 总结与关键要点
The cabinet principle transforms unwieldy case analysis into a one-line logical deduction. Its power lies in its generality across diverse topics. The key steps are: identify the items and the cabinets, check that items outnumber cabinets, and apply the appropriate form (simple or generalised). Edexcel questions reward clear statement of the principle followed by a concise application. Mastering this principle builds mathematical proof skills that extend far beyond the exam.
柜子原理将繁琐的状况分析转化为一步逻辑推演。它的力量在于跨主题的普适性。关键步骤是：明确物品和柜子，检验物品数量超过柜子数量，然后应用恰当的形式（简单或广义）。Edexcel 的考题奖赏清晰地陈述原理并接着给出简明应用的做法。掌握这个原理能培养远远超出考试的数学证明能力。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图