Combinations & Permutations: Unlocking the Number 120 | 组合与排列：解锁120的奥秘

📚 Combinations & Permutations: Unlocking the Number 120 | 组合与排列：解锁120的奥秘

Counting is one of the most intuitive yet powerful tools in mathematics. In A-Level Mathematics, the study of permutations and combinations forms the backbone of probability and statistics. Today we focus on the number 120, which appears repeatedly in counting problems: 5! = 120, 6P3 = 120, 10C3 = 120. Understanding how these values arise will unlock a deep understanding of the entire topic.

计数是数学中最直观却又最强大的工具之一。在 A-Level 数学中，排列与组合的学习构成了概率与统计的基石。今天我们将聚焦于数字 120，它频繁出现在各类计数问题中：5! = 120，6P3 = 120，10C3 = 120。理解这些数值是如何产生的，将帮助你深刻掌握整个专题。

1. The Fundamental Principle of Counting | 基本计数原理

The multiplication rule states: if one event can occur in m ways and a second event can occur in n ways, then the two events together can occur in m × n ways. For example, choosing one of 4 shirts and one of 5 pairs of trousers gives 4 × 5 = 20 outfits.

乘法原理指出：如果一件事有 m 种发生方式，另一件事有 n 种发生方式，那么两件事同时发生共有 m × n 种方式。例如，从 4 件衬衫和 5 条裤子中各选一件，共有 4 × 5 = 20 种搭配。

This principle extends to any number of events. For instance, a 3-digit number using digits 1-9 without repetition has 9 × 8 × 7 = 504 possibilities. The multiplication rule is the foundation upon which permutations and combinations are built.

这个原理可以推广到任意多个事件。例如，用 1-9 的数字组成一个没有重复数字的三位数，共有 9 × 8 × 7 = 504 种可能。乘法原理是排列和组合的基础。

2. Factorials and the Number 120 | 阶乘与数字120

The factorial of a positive integer n, written as n!, is the product of all positive integers from 1 to n. The number 120 emerges naturally as 5! because 5 × 4 × 3 × 2 × 1 = 120. It also represents the number of ways to arrange 5 distinct objects in a line.

正整数 n 的阶乘记作 n!，它表示从 1 到 n 的所有正整数的乘积。数字 120 自然地来自 5!，因为 5 × 4 × 3 × 2 × 1 = 120。它也代表 5 个不同物体排成一排的排列方式总数。

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

By definition, 0! = 1. This convention ensures that formulas for nPr and nCr work correctly when all objects are selected. For Edexcel, you must be able to evaluate factorials for n up to at least 10, and simplify expressions with factorials.

根据定义，0! = 1。这一约定确保 nPr 和 nCr 的公式在选取全部对象时依然正确。对于 Edexcel 考试，你需要能够计算 n 至少到 10 的阶乘，并化简包含阶乘的表达式。

3. Permutations: Order Matters | 排列：顺序重要

A permutation is an arrangement of objects where order is important. For example, the ways to arrange the letters A, B, C are ABC, ACB, BAC, BCA, CAB, CBA — that is 3! = 6 distinct arrangements. If we permute r objects chosen from n distinct objects, we use the notation nPr.

排列是指对象的安排，其中顺序是重要的。例如，排列字母 A、B、C 的方式有 ABC、ACB、BAC、BCA、CAB、CBA——共有 3! = 6 种不同的排列。如果从 n 个不同对象中选取 r 个进行排列，我们使用记号 nPr。

The formula for permutations is:

nPr = n! ⁄ (n − r)!

This formula counts the number of ways to select r objects from n and then arrange them in order. For example, 6P3 = 6! ⁄ 3! = 720 ⁄ 6 = 120, which is exactly the number of ways to choose 3 people from 6 and line them up.

这个公式计算从 n 个对象中选取 r 个并排成顺序的方式数。例如，6P3 = 6! ⁄ 3! = 720 ⁄ 6 = 120，这正是从 6 个人中选 3 个人并让他们排成一队的方式数。

4. Combinations: Order Does Not Matter | 组合：顺序不重要

In contrast, a combination is a selection of objects where order is irrelevant. Choosing 3 players from a team of 6 — the order of selection does not matter; the result is just the set of chosen players. The number of ways to choose r objects from n distinct objects is denoted nCr.

相比之下，组合是指不考虑顺序地选取对象。从 6 名队员中选出 3 名上场——选择的顺序无关紧要，结果只是被选中的队员集合。从 n 个不同对象中选取 r 个的方式数记作 nCr。

The formula for combinations is:

nCr = n! ⁄ [r! × (n − r)!]

Every combination gives rise to r! different permutations. This is why the permutation formula is divided by r! to obtain the combination formula. For example, choosing 3 objects from 6 gives 6C3 = 20, not 120, because each group of 3 can be ordered in 3! = 6 ways.

每一个组合对应 r! 种不同的排列。这就是为什么排列公式要除以 r! 才能得到组合公式。例如，从 6 个对象中选取 3 个得到 6C3 = 20，而不是 120，因为每组 3 个对象可以有 3! = 6 种排列顺序。

5. The Key Formulas: nPr and nCr | 关键公式：nPr 与 nCr

You must be able to switch between the two formulas and understand their relationship. Notice that nCr = nPr ⁄ r!. This is a powerful link: arranging r selected objects requires multiplying by r!.

你必须能够在两个公式之间切换并理解它们的关系。注意 nCr = nPr ⁄ r!。这是一个强有力的联系：将选出的 r 个对象排列需要乘以 r!。

nPr = nCr × r!

On the Edexcel calculator, the functions are usually denoted nPr and nCr. However, you must still be able to compute them manually using factorial expressions, especially when simplifying algebraic terms such as nC3 or (n+1)P2.

在 Edexcel 允许的计算器上，通常有 nPr 和 nCr 功能。但你仍然需要掌握用手算的阶乘表达式来计算，特别是化简像 nC3 或 (n+1)P2 这样的代数项。

Example: Simplify 10C3 using factorial notation. 10C3 = 10! ⁄ (3! × 7!) = (10 × 9 × 8) ⁄ (3 × 2 × 1) = 720 ⁄ 6 = 120. The long factorials cancel nicely.

例如：用阶乘记号化简 10C3。10C3 = 10! ⁄ (3! × 7!) = (10 × 9 × 8) ⁄ (3 × 2 × 1) = 720 ⁄ 6 = 120。冗长的阶乘可以很好地约去。

6. Why 10C3 = 120? A Step-by-Step Calculation | 为什么10C3 = 120？逐步计算

The number 120 appears in combination problems more often than you might think. 10C3 equals 120 because we are choosing 3 objects from 10 distinct objects without regard to order. Let’s break it down using the formula.

数字 120 出现在组合问题中的频率比你想象的要高。10C3 等于 120，是因为我们从 10 个不同对象中不考虑顺序地选取 3 个。让我们用公式来逐步分解。

Start with 10C3 = 10! ⁄ (3! × (10 − 3)!) = 10! ⁄ (3! × 7!). Now expand the numerator as 10 × 9 × 8 × 7! and cancel the 7!:

首先，10C3 = 10! ⁄ (3! × (10 − 3)!) = 10! ⁄ (3! × 7!)。将分子展开为 10 × 9 × 8 × 7!，并与分母中的 7! 约去：

(10 × 9 × 8) ⁄ (3 × 2 × 1) = 720 ⁄ 6 = 120

Another way to see it: the number of ordered triples from 10 is 10P3 = 10 × 9 × 8 = 720. Since each unordered set of 3 corresponds to 3! = 6 ordered arrangements, we divide 720 by 6 to get 120.

另一种理解方式：从 10 个对象中形成有序三元组共有 10P3 = 10 × 9 × 8 = 720 种。因为每一个无序三元组对应 3! = 6 种有序排列，所以将 720 除以 6 得到 120。

7. When 120 Appears: 6P3 and 5! | 120出现的时刻：6P3 与 5!

Let’s compare three situations that all give the answer 120. First, 5! counts the number of ways to arrange 5 different objects in a line. For example, arranging 5 books on a shelf gives 120 orders.

我们来比较三种答案都为 120 的情形。首先，5! 表示将 5 个不同物体排成一行的方法数。例如，将 5 本书摆上书架共有 120 种顺序。

Second, 6P3 counts the number of ways to select and arrange 3 objects from 6. For instance, awarding gold, silver, and bronze medals to 3 out of 6 athletes gives 6 × 5 × 4 = 120 possible medal distributions.

其次，6P3 表示从 6 个物体中选取并排列 3 个的方法数。例如，从 6 名运动员中颁发金、银、铜牌给其中 3 人，共有 6 × 5 × 4 = 120 种可能的奖牌归属。

Third, 10C3 counts the number of ways to choose 3 items from a menu of 10. If you can choose any 3 starters from 10, there are 120 different combinations — the order on the plate doesn’t matter. These examples highlight the difference between arranging and selecting.

第三，10C3 表示从 10 道前菜中选择 3 道的方法数。如果你可以从 10 道前菜中任选 3 道，就有 120 种不同的组合——盘子上的顺序并不重要。这些例子突出显示了排列与选择之间的区别。

8. Problem-Solving Strategy for A-Level | A-Level 解题策略

When facing a counting problem, first ask: Does order matter? If yes, use permutations. If no, use combinations. Then check whether repetition is allowed. In A-Level Edexcel, most questions involve distinct objects without repetition, but you must still be alert.

面对计数问题时，首先问：顺序重要吗？如果重要，用排列；如果不重要，用组合。然后检查是否允许重复。在 Edexcel 的 A-Level 中，大多数问题涉及没有重复的不同对象，但你必须保持警觉。

Next, identify the values of n and r. For example, “choose 2 men from 8 men” gives 8C2 = 28. “Arrange 5 books from 8” gives 8P5 = 6720. Always write down the expression before simplifying to avoid arithmetic errors.

接下来，识别 n 和 r 的值。例如，“从 8 名男士中选 2 名”得到 8C2 = 28。“从 8 本书中排列 5 本”得到 8P5 = 6720。在化简之前先写出表达式，以避免计算错误。

A common trap is confusing nPr with nCr. If there is an explicit ordering (like positions, ranks, or seating arrangements), it’s a permutation. If it’s a selection group (like committees, teams, or lottery tickets), it’s a combination.

一个常见陷阱是混淆 nPr 与 nCr。如果存在明确顺序（如职位、名次或座位安排），就是排列；如果只是选择群体（如委员会、队伍或彩票），就是组合。

9. Common Mistakes and Misconceptions | 常见错误与误解

One frequent error is treating combinations as permutations, or vice versa. For example, in a committee, the order of members does not matter; in a president-and-treasurer election, it does. Another mistake is forgetting to multiply by the factorial of the number of groups when using the multiplication rule.

一个常见错误是将组合当作排列，或反过来。例如，委员会中成员顺序无所谓；而选举主席和财务主管时顺序就有影响。另一个错误是在使用乘法原理时忘记乘以分组的阶乘。

Misinterpreting “at least” or “at most” is also common. For instance, “at least 2 girls” means 2 girls or 3 girls or more, so you must add separate combination terms. In Edexcel questions, always break compound restrictions into mutually exclusive cases and sum the results.

误解“至少”或“至多”也很常见。例如，“至少 2 名女生”意味着 2 名、3 名或更多，因此你必须分别计算组合项并相加。在 Edexcel 考题中，始终将复合限制分解为互斥的情况，然后求和。

Finally, do not cancel the factorials incorrectly. For example, n! ⁄ r! is not equal to (n − r)! unless n−r = 1. Always expand or use the cancellation property step by step.

最后，不要错误地约分阶乘。例如，n! ⁄ r! 不等于 (n − r)!，除非 n−r = 1。始终逐步展开或使用约分性质。

10. Exam-Style Question: The 120 Club | 考试风格题目：120俱乐部

Let’s solve a typical Edexcel question: A student is asked to evaluate nC4 = ⋯? Suppose n is such that nC4 = 120. Find n, and also find nP2.

让我们解决一道典型的 Edexcel 题目：已知 nC4 = 120，求 n，并求 nP2。

First, write the equation: n! ⁄ (4! × (n − 4)!) = 120. The left side simplifies to n(n−1)(n−2)(n−3) ⁄ 24. Setting this equal to 120 gives n(n−1)(n−2)(n−3) = 2880.

首先，写出方程：n! ⁄ (4! × (n − 4)!) = 120。左边化简为 n(n−1)(n−2)(n−3) ⁄ 24。令其等于 120，得 n(n−1)(n−2)(n−3) = 2880。

Now test possible integer values. Since 10 × 9 × 8 × 7 = 5040, which is too large, and 9 × 8 × 7 × 6 = 3024, which is also larger than 2880, while 8 × 7 × 6 × 5 = 1680, too small. Wait — 9 × 8 × 7 × 6 = 3024; 2880 is not exactly this. Perhaps nC4 = 120 when n = 10? Let’s check: 10C4 = 210, not 120. So the equation must be different.

现在尝试可能的整数值。由于 10 × 9 × 8 × 7 = 5040 太大，9 × 8 × 7 × 6 = 3024 也大于 2880，而 8 × 7 × 6 × 5 = 1680 太小。等等——9 × 8 × 7 × 6 = 3024；2880 并不等于这个值。也许 nC4 = 120 时 n = 8？检查：8C4 = 70。没有直接解。

But we know that 10C3 = 120, so if the question says nC3 = 120, then n = 10. This highlights the importance of checking the value of r in a problem. If we instead have nC3 = 120, then n(n−1)(n−2) = 720, and n = 10 gives 10 × 9 × 8 = 720. So n = 10. Then nP2 = 10P2 = 10 × 9 = 90.

但我们知道 10C3 = 120，所以如果题目说 nC3 = 120，那么 n = 10。这凸显了检查题目中 r 值的重要性。如果我们改为 nC3 = 120，则 n(n−1)(n−2) = 720，而 n = 10 时 10 × 9 × 8 = 720。因此 n = 10。于是 nP2 = 10P2 = 10 × 9 = 90。

11. Practice Problems to Master 120 | 掌握120的练习题目

Try these exercises on your own. First, compute 7P3 and 7C3 and compare the results. Second, show that 6C2 = 6C4. Third, find the number of ways to choose 3 representatives from a group of 10 students (you should get 120). Fourth, find the number of ways to arrange 3 selected representatives in a row (you should get 720, which is 120 × 6).

请自行尝试以下练习。第一，计算 7P3 和 7C3 并比较结果。第二，证明 6C2 = 6C4。第三，求从 10 名学生中选 3 名代表的方式数（你应该得到 120）。第四，求将所选 3 名代表排成一排的方式数（你应该得到 720，即 120 × 6）。

For more challenge, solve: How many different committees of 3 people can be formed from 6 men and 4 women if the committee must include at least one woman? Use 10C3 − 6C3 = 120 − 20 = 100. This is a classic Edexcel style problem that uses the complement method.

更有挑战性的题目：从 6 名男性和 4 名女性中组成一个 3 人委员会，要求至少包含一名女性，共有多少种方式？使用 10C3 − 6C3 = 120 − 20 = 100。这是一道经典的 Edexcel 风格题目，采用补集法。

12. Conclusion: The Power of 120 | 结语：120的力量

The number 120 is more than a random value; it connects many counting concepts. Whether it appears as 5!, 6P3, or 10C3, understanding how it is derived reinforces your ability to solve permutation and combination problems with confidence.

数字 120 不仅仅是一个随意的数值；它连接了许多计数概念。无论它以 5!、6P3 还是 10C3 出现，理解它的推导过程都能增强你自信地解决排列与组合问题的能力。

Remember the core principles: factorials count arrangements, permutations count ordered arrangements, and combinations count unordered selections. Master these tools, and you will not only crack every counting question in the Edexcel exam — you will also see the beauty behind numbers like 120.

请记住核心原则：阶乘计数有序整体的排列，排列计数有序选取，组合计数无序选取。掌握这些工具，你不仅能解决 Edexcel 考试中的所有计数问题，还能看到像 120 这样的数字背后的美。

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Combinations & Permutations: Unlocking the Number 120 | 组合与排列：解锁120的奥秘