📚 Counting and the Binomial Expansion | 计数与二项式展开
Counting and the binomial expansion are fundamental topics in IB Mathematics, linking combinatorics and algebra. Mastering how to count arrangements and selections underpins probability calculations and the expansion of expressions like (a + b)ⁿ. This article covers the core principles from basic counting to the binomial theorem, including permutations, combinations, Pascal's triangle, and techniques for extracting specific terms, with a brief look at the extension to rational exponents found in the HL syllabus.
计数与二项式展开是 IB 数学中的基础主题，将组合学与代数联系起来。掌握如何计算排列与组合不仅能支撑概率计算，也是展开 (a + b)ⁿ 等表达式的关键。本文涵盖了从基本计数原理到二项式定理的核心内容，包括排列、组合、帕斯卡三角和求特定项的方法，并简要介绍 HL 课程中对有理指数的扩展。
1. Fundamental Counting Principle | 基本计数原理
× n ways. This multiplicative rule extends naturally to any number of successive choices.
如果一个事件有 m 种发生方式，另一个独立事件有 n 种发生方式，那么这两个事件先后发生共有 m × n 种方式。这一乘法规则可以自然地推广到任意多个连续选择的情况。
If you also choose between 2 hats, the total becomes 3 × 2 × 2 = 12 distinct combinations.
例如，假设你有 3 件不同的衬衫和 2 条不同的裤子，可能的搭配总数就是 3 × 2 = 6。如果再在 2 顶帽子中选择，总数就变成 3 × 2 × 2 = 12 种不同的组合。
2. Factorial Notation | 阶乘记号
The factorial of a positive integer n, denoted n!, is the product of all positive integers from 1 to n: n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1. By convention, 0! = 1, which keeps formulas consistent when n = 0 appears in combinations and permutations.
正整数 n 的阶乘，记作 n!，是 1 到 n 所有正整数的乘积：n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1。按照约定，0! = 1，这样当 n = 0 出现在组合和排列公式中时能保持一致性。
Factorials grow extremely fast: 5! = 120, 10! 3 628 800. They provide the building blocks for counting arrangements where order matters.
阶乘的增长极其迅速：5! = 120，10! = 3 628 800。它们为考虑顺序的计数问题提供了基石。
3. Permutations of Distinct Objects | 不同对象的排列
A permutation is an ordered arrangement of objects. The number of ways to arrange n distinct objects in a line is n!. This follows directly from the fundamental counting principle: n choices for the first position, (n − 1) for the second, and so on.
排列是指对象的有序安排。将 n 个不同的对象排成一列的方法数是 n!。这直接由基本计数原理得到：第一个位置有 n 种选择，第二个位置有 (n − 1) 种，以此类推。
objects taken from a set of n distinct objects, the number of permutations is P(n, r) = n! / (n − r)!. For instance, the number of ways to choose a president, vice‑president and secretary from 10 students is P(10, 3) = 10 × 9 × 8 = 720.
当我们只从 n 个不同对象中选取 r 个进行排列时，排列数为 P(n, r) = n! / (n − r)!。比如，从 10 名学生中选出一位主席、一位副主席和一位秘书的选法数为 P(10, 3) = 10 × 9 × 8 = 720。
4. Permutations with Restrictions | 有限制条件的排列
Many IB problems impose constraints, such as certain items must be adjacent or separated. To handle items that must stay together, treat the group as a single block, permute the block internally, and then permute the whole set.
许多 IB 题目会施加约束，例如某些元素必须相邻或彼此分开。处理必须相邻的元素时，可以把这一组看作一个整体，先在组内进行排列，再对整个集合进行排列。
ng the letters A, B, C, D so that A and B are adjacent, treat AB as one block. There are 3! ways to arrange the block AB, C, and D, and 2! ways to arrange A and B within the block, giving 3! × 2! ed by subtracting the arrangements where objects are together from the unrestricted total.
例如，排列字母 A、B、C、D 且要求 A 与 B 相邻，可以把 AB 看作一个整体。将整体 AB、C、D 进行排列有 3! 种方式，而 AB 内部有 2! 种排列方式，于是共有 3! × 2! = 12 种排法。涉及“相互分离”的限制通常通过从无限制的总排列数中减去对象相邻的排列数来解决。
5. Combinations | 组合
not matter. The number of ways to choose r objects from n distinct objects is C(n, r) = n! / [r! (n − r)!]. This is often read as 'n choose r' and also written as nCr.
组合是指不考虑顺序的对象选取。从 n 个不同对象中选取 r 个的方法数是 C(n, r) = n! / [r! (n − r)!]，常读作“n 选 r”，也写作 nCr。
the roles. The relationship between permutations and combinations is P(n, r) = C(n, r) × r!, since each selection of r items can be arranged in r! ways.
例如，从 10 人中选出 3 人组成一个委员会（不区分职务）有 C(10, 3) = 120 种可能的组成方式。排列与组合的关系是 P(n, r) = C(n, r) × r!，因为每一组选出的 r 个元素都可以被排列成 r! 种顺序。
6. The Binomial Theorem for Positive Integer Exponents | 正整数指数的二项式定理
For any positive integer n, the expansion of (a + b)ⁿ is a sum of terms involving powers of a and b with coefficients given by combinations. The theorem states:
对于任意正整数 n，(a + b)ⁿ 的展开是由 a 和 b 的幂次与组合数系数构成的项之和。定理表述为：
(a + b)ⁿ = Σ (from r = 0 to n) C(n, r) aⁿ⁻ʳ bʳ
C(4,0)x⁴ + C(4,1)x³·2 + C(4,2)x²·2² + C(4,3)x·2³ + C(4,4)·2⁴ = x⁴ + 8x³ + 24x² + 32x + 16.
这意味着展开式包含 (n + 1) 项。例如， C(4,0)x⁴ + C(4,1)x³·2 + C(4,2)x²·2² + C(4,3)x·2³ + C(4,4)·2⁴ = x⁴ + 8x³ + 24x² + 32x + 16。
7. Binomial Coefficients and Pascal's Triangle | 二项式系数与帕斯卡三角
Binomial coefficients C(n, r) form a symmetric pattern. They can be arranged in Pascal' C(n − 1, r − 1) + C(n − 1, r). The triangle begins with row 0: 1; row 1: 1 1; row 2: 1 2 1; row 3: 1 3 3 1, and so on.
二项式系数 C(n, r) 构成对称的模式。它们可以排成帕斯卡三角，其中每个数都是其上方两数之和：C(n, r) = C(n − 1, r − 1) + C(n − 1, r)。三角形从第 0 行开始：1；第 1 行：1 1；第 2 行：1 2 1；第 3 行：1 3 3 1，依此类推。
Pascal' like symmetry: C(n, r) = C(n, n − r). It also reveals that the sum of the numbers in the n‑th row is 2ⁿ.
帕斯卡三角为较小的 n 提供了快速获取系数的方法，并展示出对称性 C(n, r) = C(n, n − r) 等性质。它还揭示了第 n 行数字之和为 2ⁿ。
8. General Term of a Binomial Expansion | 二项展开式的通项
It is often more useful to work with the general term rather than the full expansion. In (a + b)ⁿ, the term containing bʳ is the (r + 1)‑th term and is given by:
在很多情况下，使用通项比写出完整展开式更为实用。在 (a + b)ⁿ 中，含有 bʳ 的项是第 (r + 1) 项，其表达式为：
Tᵣ₊₁ = C(n, r) aⁿ⁻ʳ bʳ
f the binomial involves subtraction, say (a − b)ⁿ, the sign alternates: Tᵣ₊₁ = C(n, r) aⁿ⁻ʳ (−b)ʳ = (−1)ʳ C(n, r) aⁿ⁻ʳ bʳ.
这里 r 从 0（第一项）取到 n。如果二项式包含减法，例如 (a − b)ⁿ，则符号交替出现：Tᵣ₊₁ = C(n, r) aⁿ⁻ʳ (−b)ʳ = (−1)ʳ C(n, r) aⁿ⁻ʳ bʳ。
9. Expanding (a + b)ⁿ | 展开 (a + b)ⁿ
a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵. Notice the coefficients 1, 5, 10, 10, 5, 1 match Pascal's triangle row 5.
要系统地展开 (a + b)ⁿ，可以按 b 的升幂顺序列出各项。对于 n = 5，展开式为 (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵。注意系数 1, 5, 10, 10, 5, 1 恰好对应帕斯卡三角的第 5 行。
y)³ = A³ + 3A²B + 3AB² + B³ = 8x³ − 36x²y + 54xy² − 27y³. Always simplify each term fully.
如果表达式中含有系数，例如 (2x − 3y)³，可先令 A = 2x，B = −3y。那么 (2x − 3y)³ = A³ + 3A²B + 3AB² + B³ = 8x³ − 36x²y + 54xy² − 27y³。务必对每一项进行彻底化简。
10. Finding a Specific Term in an Expansion | 求展开式中的特定项
Exam questions often ask for a term independent of x (constant term) or the coefficient of x raised to a given power. Set up the general term Tᵣ₊₁, simplify the powers of x, equate the index to the desired exponent, and solve for r.
考试中常见的问题包括求与 x 无关的常数项，或求 x 的某次幂的系数。写出通项 Tᵣ₊₁，化简 x 的指数，令其等于所需的幂指数，再解出 r。
xample, find the constant term in (2x² − 1/x)⁶. The general term is Tᵣ₊₁ = C(6, r) (2x²)⁶⁻ʳ (−1/x)ʳ = C(6, r) 2⁶⁻ʳ (−1)ʳ x¹²⁻²ʳ⁻ʳ = C(6, r) 2⁶⁻ʳ (−1)ʳ x¹²⁻³ʳ. For the constant term, the exponent of x must be 0, so 12 − 3r = 0 ⇒ r = 4. Substituting r = 4 gives the term C(6, 4) × 2² × (−1)⁴ = 15 × 4 × 1 = 60.
例如，求 (2x² − 1/x)⁶ 的常数项。通项为 Tᵣ₊₁ = C(6, r) (2x²)⁶⁻ʳ (−1/x)ʳ = C(6, r) 2⁶⁻ʳ (−1)ʳ x¹²⁻²ʳ⁻ʳ = C(6, r) 2⁶⁻ʳ (−1)ʳ x¹²⁻³ʳ。对于常数项，x 的指数必须为 0，因此 12 − 3r = 0 ⇒ r = 4。代入 r = 4 得 C(6, 4) × 2² × (−1)⁴ = 15 × 4 × 1 = 60。
11. Sum of Binomial Coefficients and Applications | 二项式系数之和及应用
theorem: (1 + 1)ⁿ = Σ C(n, r) = 2ⁿ. This confirms that the total number of subsets of an n‑element set is 2ⁿ, each choice of r elements corresponding to a combination.
在二项式定理中令 a = b = 1，会得到一个经典结果：(1 + 1)ⁿ = Σ C(n, r) = 2ⁿ。这印证了 n 元集合的所有子集个数为 2ⁿ，每个选取 r 个元素的组合都对应一个子集。
By setting a = 1, b = −1, we get 0 = Σ (−1)ʳ C(n, r), which shows that the sum of coefficients of even‑indexed terms equals the sum of odd‑ uch properties are useful in problems involving alternating sums of binomial coefficients.
令 a = 1，b = −1，可得 0 = Σ (−1)ʳ C(n, r)，这表明偶数项系数之和等于奇数项系数之和。这类性质在涉及二项式系数交错求和的问题中非常有用。
12. Extension: Binomial Series for Rational Exponents (HL) | 扩展：有理指数的二项式级数 (HL)
For IB HL Mathematics, the binomial theorem is extended to exponents that are not positive integers. If n is a rational number and |x| < 1, then (1 + x)ⁿ can be expanded as an infinite series:
在 IB HL 数学中，二项式定理被推广到指数不是正整数的情形。如果 n 是有理数且 |x| < 1，则 (1 + x)ⁿ 可以展开为一个无穷级数：
(1 + x)ⁿ = 1 + nx + [n(n − 1)/2!] x² + [n(n − 1)(n − 2)/3!] x³ + …
Here the coefficient of xʳ is n(n − 1)…(n − r + 1) / r!, which generalises C(n, r). The expansion is valid for |x| < 1. This formula allows approximations of roots and reciprocals: for example, √1 + x ≈ 1 + x/2 − x²/8 for small x.
这里 xʳ 的系数为 n(n − 1)…(n − r + 1) / r!，它正是 C(n, r) 的推广。该展开在 |x| < 1 时成立。此公式可用于根式和倒数的逼近：例如当 x 很小时，√1 + x ≈ 1 + x/2 − x²/8。
While the series is infinite, in exam contexts students are usually asked to expand up to a specified number of terms or to state the range of validity. Always check |x| < 1 before applying the infinite expansion.
虽然级数是无穷的，但在考试中通常只要求学生展开到指定项数或注明成立范围。在应用无穷展开之前，务必检查是否满足 |x| < 1
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