Factorial Notation | 阶乘记法

📚 Factorial Notation | 阶乘记法

Factorial notation is a compact way of writing the product of consecutive positive integers. It appears throughout Edexcel A-Level Mathematics, especially in permutations, combinations and binomial expansions. Understanding how to simplify factorial expressions quickly is a core skill for both Pure and Statistics units.

阶乘记法是一种简洁地表示连续正整数乘积的方法。在 Edexcel A-Level 数学中，它广泛出现在排列、组合和二项式展开等内容中。快速化简阶乘表达式是纯数学和统计单元的核心技能。

This guide covers the definition, the special value 0! ification techniques, and the way factorials power permutations, combinations, binomial coefficients, probability and statistics questions.

本文涵盖阶乘的定义、特殊值 0! = 1、递推关系、化简技巧，以及阶乘在排列、组合、二项式系数、概率与统计题目中的作用。

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1. Definition of Factorial Notation | 阶乘记法的定义

For a positive integer n, n! (read as "n factorial") is the product of all positive integers from n down to 1. The exclamation mark is part of the mathematical symbol, not punctuation, so it must never be dropped or replaced.

对于正整数 n，n!（读作"n 的阶乘"）表示从 n 一直乘到 1 的所有正整数的乘积。感叹号是数学符号的一部分，不是标点，因此不能省略或用别的符号代替。

n! = n × (n − 1) × (n − 2) × ⋯ × 3 × 2 × 1

great deal of writing. Instead of writing 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, we simply write 8! matics the factorial is defined only for non-negative integers, so expressions such as 2.5! or (−3)! do not appear in the specification.

这种紧凑的记法可以节省大量书写。与其写成 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1，我们只需写成 8!。在 A-Level 数学中，阶乘只对非负整数有定义，因此像 2.5! 或 (−3)! 这样的表达式不会出现在考纲中。

For example, 5! = 5 × 4 × 3 × 2 × 1 = 120 and 3! = 3 × 2 × 1 = 6. A very common slip is to read 5! × 4; in fact it is a chain of factors that decreases by one each time until it reaches 1.

例如，5! = 5 × 4 × 3 × 2 × 1 = 120，3! = 3 × 2 × 1 = 6。一个非常常见的错误是把 5! 理解成 5 × 5 或 5 × 4；实际上它是一串每次递减 1 的因子，一直乘到 1 为止。

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2. Values of Small Factorials | 较小阶乘的数值

You should know the small factorial values instantly, because they appear repeatedly in counting arguments and in binomial expansions. The table below lists the standard values.

你应当熟记较小的阶乘值，因为它们在计数推理和二项式展开中反复出现。下表列出了标准数值。

n	n!	Value | 数值
0	0!	1
1	1!	1
2	2!	2
3	3!	6
4	4!	24
5	5!	120
6	6!	720
7	7!	5040
8	8!	40320
9	9!	362880
10	10!	3628800

Values grow explosively: 10! is already greater than three million, and 20! owth, exam questions rarely ask you to multiply out large factorials; instead they ask you to simplify, cancel or compare them.

阶乘数值增长极快：10! 已经超过三百万，而 20! ≈ 2.43 × 10¹⁸。由于增长如此迅速，考试题很少要求你把大阶乘乘开，而是要求你化简、约分或比较它们。

TI-Nspire CX II or a Casio ClassWiz, the factorial is available directly from the probability or catalogue menu. Note that most calculators overflow above 69!, since 69! lay limit, so an exact answer may need to be written in prime-factorised or algebraic form.

在 TI-Nspire CX II 或 Casio ClassWiz 等图形计算器上，可以直接从概率或目录菜单中调用阶乘功能。要注意，多数计算器在超过 69! 时会溢出，因为 69! ≈ 1.71 × 10⁹⁸ 已接近 10¹⁰⁰ 的显示上限，因此精确答案有时需要写成质因数分解形式或代数形式。

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3. The Special Case 0! = 1 | 特殊情况 0! = 1

By definition, 0! nvention that keeps every formula in combinatorics consistent, so you should treat it as a fact to be quoted rather than proved.

根据定义，0! = 1。这是一个约定，但它能让组合数学中的所有公式保持一致，因此你应当把它当作可以直接引用的事实，而不需要在考试中证明它。

0! = 1

The recursive definition gives a neat justification. Since n! = n × (n − 1)!, putting n = 1 gives 1! = 1 × 0!. Because 1! = 1, we must have 0! = 1. This is an example of the "empty product" convention: multiplying no numbers together leaves the multiplicative identity 1.

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递推定义给出了一个简洁的理由。由于 n! = n × (n − 1)!，令 n = 1 得 1! = 1 × 0!。因为 1! = 1，所以必须有 0! = 1。这体现了"空乘积"约定：不乘任何数时，结果就是乘法单位元 1。

re is also a counting reason. The number of ways of arranging r objects chosen from n is n! ÷ (n − r)!, and the number of ways of choosing r objects from n is n! ÷ (r!(n − r)!). If you choose all n objects, or choose none of them, the formula only works when 0! = 1. There is exactly one way to arrange nothing at all.

还有一个计数上的理由。从 n 个对象中选出 r 个并排列的方式数为 n! ÷ (n − r)!，而从中选出 r 个的方式数为 n! ÷ (r!(n − r)!)。当你取全部 n 个或一个都不取时，只有 0! = 1 才能让公式成立。什么都不排，恰好只有一种方式。

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4. The Recursive Relationship n! = n × (n − 1)! | 递推关系 n! = n × (n − 1)!

The single most useful identity in factorial algebra is that a factorial can be peeled back one factor at a time. Removing the largest factor leaves the factorial of the previous integer.

阶乘代数中最有用的恒等式是：阶乘可以一次剥掉一个因子。去掉最大的因子后，剩下的就是前一个整数的阶乘。

n! = n × (n − 1)!

So 10! = 10 × 9!, 7! = 7 × 6!, and in general n! = n × (n − 1) × (n − 2)! and so on. You can extend upwards too: (n + 1)! = (n + 1) × n!.

因此 10! = 10 × 9!，7! = 7 × 6!，一般地 n! = n × (n − 1) × (n − 2)!，依此类推。你也可以向上扩展：(n + 1)! = (n + 1) × n!。

This identity is the engine behind almost every simplification question. Whenever you see a factorial quotient with algebraic expressions such as (n + 2)! or (2n − 1)!, rewrite the larger factorial as a product of the smaller one and a few extra factors, then cancel.

这个恒等式几乎是所有化简题的"发动机"。只要看到含 (n + 2)! 或 (2n − 1)! 这类代数式的阶乘商，就把较大的阶乘改写成较小阶乘乘以若干额外因子的形式，然后约分。

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5. Simplifying Ratios of Factorials | 阶乘之比的化简

Factorial quotients are simplified by writing out only the extra factors. The general pattern is that n! ÷ (n − r)! is the product of the r largest factors from n downwards.

阶乘之商的化简方法是只写出多出来的因子。一般规律是：n! ÷ (n − r)! 等于从 n 开始往下数 r 个最大因子的乘积。

n! ÷ (n − r)! = n × (n − 1) × (n − 2) × ⋯ × (n − r + 1)

For example, n! ÷ (n − 2)! = n(n − 1), and (n + 2)! ÷ n! = (n + 2)(n + 1). Similarly (2n)! ÷ (2n − 1)! = 2n, because only the single factor 2n is left over after cancellation.

例如，n! ÷ (n − 2)! = n(n − 1)，而 (n + 2)! ÷ n! = (n + 2)(n + 1)。同样，(2n)! ÷ (2n − 1)! = 2n，因为约分之后只剩下单个因子 2n。

Here is an exam-style equation. Solve n! ÷ (n − 2)! = 42 for the positive integer n.

下面是一道考试风格的方程题。对正整数 n，解方程 n! ÷ (n − 2)! = 42。

n(n − 1) = 42 → n² − n − 42 = 0 → (n − 7)(n + 6) = 0

t the negative root explicitly, because many mark schemes award a mark for that statement.

所以 n = 7 或 n = −6。由于 n 必须是正整数，故 n = 7。务必明确舍去负根，因为很多评分标准会为这一步单独给分。

A second worked example: (n + 3)! ÷ n! = (n +

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