📚 The Binomial Coefficient: Definition, Formula and Applications | 二项式系数：定义、公式与应用
In IB Mathematics, the binomial coefficient is a core idea that links combinatorics, algebra, and probability. It tells us how many ways we can choose r items from n distinct items without regard to order, and it provides the multipliers in the expansion of (a + b)ⁿ. Whether you are solving counting problems, expanding brackets, or working with the binomial distribution, a solid understanding of binomial coefficients is essential.
在 IB 数学中，二项式系数是连接组合学、代数和概率的核心概念。它告诉我们从 n 个不同物品中无序选取 r 个物品有多少种方法，并给出了 (a + b)ⁿ 展开式中的乘数。无论你是解决计数问题、展开括号，还是处理二项分布，透彻理解二项式系数都是必不可少的。
This article builds the topic from the ground up: notation, the factorial formula, Pascal's triangle, the addition identity, the binomial theorem, general-term techniques, counting applications and the binomial distribution. Each idea is followed by worked examples and exam-style warnings so that you can move from definition to confident calculation.
本文将从头构建这一主题：符号表示、阶乘公式、杨辉三角、加法恒等式、二项式定理、通项技巧、计数应用以及二项分布。每个知识点后都配有例题与考试提醒，帮助你从定义直接走向熟练计算。
1. Definition and Notation | 定义与符号
A binomial coefficient counts the number of combinations of r objects chosen from a set of n distinct objects. It is commonly written as C(n, r), pronounced 'n choose r'. For specific numbers we may write ⁵C₂ to mean '5 choose 2', and in general the notation can appear as ⁿCᵣ or as a column-vector form with n above r and no fraction bar.
二项式系数计算从 n 个不同对象中选取 r 个对象的组合数。它通常写作 C(n, r)，读作 "n 选 r"。对于具体数字，我们可以写成 ⁵C₂ 表示 "5 选 2"，一般形式也可记为 ⁿCᵣ，或用 n 在上、r 在下、不带分数线的列向量形式表示。
The key word in the definition is 'without regard to order': the group {Amy, Ben, Cara} is the same selection as {Cara, Amy, Ben}. If order mattered, we would be counting permutations instead, and the answer would be larger by a factor of r!.
定义中的关键词是 "不考虑顺序"：组合 {Amy, Ben, Cara} 与 {Cara, Amy, Ben} 是同一种选法。若顺序重要，我们统计的就是排列，答案会大出 r! 倍。
e C(n, 0) = 1 and C(n, n) = 1, reflecting that there is exactly one way to choose no objects or to choose all of them.
根据定义，如果 r 大于 n 或 r 为负数，二项式系数等于 0，因为这样的选法不存在。此外，对任意 n ≥ 0，都有 C(n, 0) = 1 和 C(n, n) = 1，这说明不选任何对象或全选都只有一种方式。
- Order does not matter in C(n, r); it does matter in P(n, r).
- C(n, r) is always a whole number, even though its formula is a fraction.
- Valid inputs satisfy 0 ≤ r ≤ n when n is a non-negative integer.
- 在 C(n, r) 中顺序不重要；在 P(n, r) 中顺序重要。
- C(n, r) 永远是整数，尽管它的公式写成分数形式。
- 当 n 为非负整数时，有效输入满足 0 ≤ r ≤ n。
2. Special Values and Symmetry | 特殊值与对称性
e item from n items can be done in n ways. The third is C(n, 2) = n(n − 1) / 2, the standard 'handshake' count for n people.
三个特殊值在考试中反复出现。第一是 C(n, 0) = C(n, n) = 1。第二是 C(n, 1) = C(n, n − 1) = n，因为从 n 个物品中选出 1 个有 n 种方法。第三是 C(n, 2) = n(n − 1) / 2，即 n 个人握手的标准计数。
− r) is extremely useful. Choosing which r items to keep is the same as choosing which n − r items to leave behind, so both counts must be equal. In practice this means you can always replace a large r by a smaller n − r before calculating.
对称关系 C(n, r) = C(n, n − r) 极为有用。选出要保留的 r 个对象，等价于选出要留下的 n − r 个对象之外的部分，因此两个计数必然相等。实际计算中，这意味你总可以把较大的 r 换成较小的 n − r 再算。
A third family of results concerns the sum of a whole row: the sum of all binomial coefficients with a fixed n equals 2ⁿ.
第三类结论涉及整行求和：固定 n 时，所有二项式系数之和等于 2ⁿ。
C(n, 0) + C(n, 1) + C(n, 2) + … + C(n, n) = 2ⁿ
n the binomial theorem, and it explains why a set with n elements has 2ⁿ subsets in total.
这一恒等式可由二项式定理中令 a = b = 1 得到，它解释了为什么含 n 个元素的集合共有 2ⁿ 个子集。
3. Factorial Formula | 阶乘公式
formula applies for all integers with 0 ≤ r ≤ n, and it ties in with the convention that 0! = 1.
二项式系数用阶乘公式计算。该标准公式适用于所有满足 0 ≤ r ≤ n 的整数，并与 0! = 1 的约定相配合。
C(n, r) = n! ÷ (r! × (n − r)!)
For example, C(5, 2) = 5! ÷ (2! × 3!) = 120 ÷ (2 × 6) = 10. The numerator 5! is the number of ways to arrange all five items in a row; dividing by 2! removes the internal ordering of the chosen pair, and dividing by 3! removes the internal ordering of the three items not chosen.
例如，C(5, 2) = 5! ÷ (2! × 3!) = 120 ÷ (2 × 6) = 10。分子 5! 是把 5 个物品全部排成一列的排列数；除以 2! 消除被选中的两个对象内部的顺序，除以 3! 消除未被选中的三个对象内部的顺序。
When r is small, the descending-product version is faster and less error-prone, because most factorials cancel.
当 r 较小时，使用降幂连乘的形式更快、更不易出错，因为大部分阶乘会约掉。
C(n, r) = n(n − 1)(n − 2) … (n − r + 1) ÷ r!
numerator always contains exactly r factors, and the last factor is n − r + 1, not n − r.
用这一形式，C(8, 3) = (8 × 7 × 6) ÷ (3 × 2 × 1) = 336 ÷ 6 = 56。注意分子恰好有 r 个因子，最后一个因子是 n − r + 1，而不是 n − r。
4. Worked Numerical Examples | 数值计算实例
ost frequently used coefficients; learning row n = 5 and row n = 6 by heart is worthwhile.
熟练计算小数值能在长题中节省时间。下表汇总了最常用的系数，把 n = 5 和 n = 6 两行记住是值得的。
| C(4, 2) | 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 |
| C(5, 2) | 120 ÷ (2 × 6) = 10 |
| C(6, 3) | 720 ÷ (6 × 6) = 20 |
| C(8, 3) | (8 × 7 × 6) ÷ 6 = 56 |
| C(10, 2) | (10 × 9) ÷ 2 = 45 |
| C(12, 1) | 12 |
Two checks are always available. First, use symmetry: if your value of r exceeds n ÷ 2, recalculate with n − r. Second, use the row-sum identity: the entries of row n must add to 2ⁿ, so an error is easy to spot.
有两个随时可用的检验方法。第一是用对称性：如果 r 超过 n ÷ 2，就改用 n − r 重算。第二是用整行求和恒等式：第 n 行的各项之和必须等于 2ⁿ，因此错误很容易被发现。
32 = 2⁵ confirms row n = 5, and 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2⁶ confirms row n = 6.
例如，1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵ 验证了 n = 5 这一行，1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2⁶ 验证了 n = 6 这一行。
5. Pascal's Triangle | 杨辉三角
Pascal's triangle arranges the binomial coefficients in rows, where row n contains C(n, 0) through C(n, n). Each row begins and ends with 1, and every interior entry is the sum of the two entries directly above it.
杨辉三角把二项式系数按行排列，第 n 行包含 C(n, 0) 到 C(n, n)。每行以 1 开头、以 1 结尾，中间的每一项都等于其正上方两个数之和。
| n = 0 | 1 |
| n = 1 | 1 1 |
| n = 2 | 1 2 1 |
| n = 3 | 1 3 3 1 |
| n = 4 | 1 4 6 4 1 |
| n = 5 | 1 5 10 10 5 1 |
| n = 6 | 1 6 15 20 15 6 1 |
The triangle is not just a memory aid. It is the fastest route to a full expansion when n is small, and it also displays the symmetry relation visually: reading any row from left to right gives the same sequence as reading it from right to left.
杨辉三角不只是记忆工具。当 n 较小时，它是得到完整展开式最快的途径；它也把对称关系直观地展示出来：任何一行从左往右读与从右往左读得到相同数列。
Other patterns worth knowing include the natural numbers in the second diagonal, the triangular numbers 1, 3, 6, 10, 15 in the third diagonal, and the fact that the row sum doubles each time n increases by 1, which is exactly the statement that 2ⁿ doubles.
其他值得知道的规律包括：第二条对角线是自然数，第三条对角线是三角数 1, 3, 6, 10, 15；以及每增加一行其和翻倍，这正好对应 2ⁿ 的翻倍性质。
6. Pascal's Identity (Addition Formula) | 帕斯卡恒等式（加法公式）
The rule used to build the triangle is Pascal's identity, sometimes called the addition formula. It expresses a coefficient in row n in terms of two coefficients in row n − 1.
用来构造杨辉三角的规则就是帕斯卡恒等式，有时也称为加法公式。它把第 n 行的一个系数表示为第 n − 1 行两个系数之和。
C(n, r) = C(n − 1, r − 1) + C(n − 1, r)
A combinatorial proof is short and convincing. Fix one particular item, say Amy, among the n available. Every selection of r items either includes Amy, in which case the remaining r − 1 items come from the other n − 1 items, or excludes Amy, in which case all r items come from the other n − 1 items. The two cases together cover all selections, so the counts add.
组合证明简短而有力。在 n 个对象中固定一个特定对象，比如 Amy。每一种选取 r 个对象的方案，要么包含 Amy，此时其余 r − 1 个从另外 n − 1 个中选；要么不包含 Amy，此时 r 个全部从另外 n − 1 个中选。两种情况涵盖所有方案，因此计数相加。
= C(6, 3) + C(6, 4) = 20 + 15 = 35. Both results agree with direct factorial calculation.
用数值检验：C(6, 3) = C(5, 2) + C(5, 3) = 10 + 10 = 20，C(7, 4) = C(6, 3) + C(6, 4) = 20 + 15 = 35。两个结果都与直接使用阶乘公式一致。
ick pattern, for example C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) = C(7, 4), which equals 35.
反复使用该恒等式还能得到 "曲棍球杆" 规律，例如 C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) = C(7, 4) = 35。
7. The Binomial Theorem | 二项式定理
The binomial theorem states that for any positive integer n, the expansion of (a + b)ⁿ uses the binomial coefficients as multipliers. The index r runs from 0 to n, and the powers of a decrease while the powers of b increase.
二项式定理指出，对任意正整数 n，(a + b)ⁿ 的展开式以二项式系数为乘数。指标 r 从 0 取到 n，a 的幂逐项下降，b 的幂逐项上升。
(a + b)ⁿ = Σ C(n, r) aⁿ⁻ʳ bʳ (r = 0 to n)
Writing the first few terms explicitly shows the structure clearly, and the powers in every term always add to n.
把前几项写出来能清楚展示结构，而且每一项中两个幂指数之和始终等于 n。
(a + b)ⁿ = aⁿ + C(n, 1) aⁿ⁻¹ b + C(n, 2) aⁿ⁻² b² + … + bⁿ
For example, expanding (x + 2)⁴ by row n = 4 of Pascal's triangle gives the coefficients 1, 4, 6, 4, 1.
例如，用杨辉三角第 n = 4 行去展开 (x + 2)⁴，系数为 1, 4, 6, 4, 1。
(x + 2)⁴ = x⁴ + 8x³ + 24x² + 32x + 16
Each coefficient comes from C(4, r) multiplied by 2ʳ 16. Forgetting the powers of the constant 2 is the single most common error in this topic.
每个系数都由 C(4, r) 乘以 2ʳ 得到：C(4, 1) × 2 = 8，C(4, 2) × 2² = 6 × 4 = 24，C(4, 3) × 2³ = 4 × 8 = 32，C(4, 4) × 2⁴ = 16。忘记常数 2 的幂是本主题中最常见的错误。
8. Finding a Particular Term | 求特定项
The general term, counting from r = 0, is the (r + 1)th term of the expansion of (a + b)ⁿ.
大多数考题并不要求写出整个展开式，而是要求某一项或某个系数。从 r = 0 开始计数，通项是 (a + b)ⁿ 展开式中的第 (r + 1) 项。
Tᵣ₊₁ = C(n, r) aⁿ⁻ʳ bʳ
ficient of x⁵ in the expansion of (2 + 3x)⁷. Using a = 2 and b = 3x, the term containing xʳ is C(7, r) 2⁷⁻ʳ (3x)ʳ, so we need r = 5.
例题：求 (2 + 3x)⁷ 展开式中 x⁵ 的系数。取 a = 2、b = 3x，含 xʳ 的项为 C(7, r) 2⁷⁻ʳ (3x)ʳ，因此需要 r = 5。
C(7, 5) × 2² × 3⁵ = 21 × 4 × 243 = 20412
us sign must be raised to the same power as the term, and the answer stays negative because 3 is odd.
第二个含负号的例子：(1 − 2x)⁶ 中 x³ 的系数为 C(6, 3) × (−2)³ = 20 × (−8) = −160。负号也必须升到与该项相同的幂次，因为 3 是奇数，结果保持为负。
The routine for these questions is always the same: write the general term, set the exponent of x equal to the required value, solve for r, then substitute r back and evaluate the product. Do not skip the middle step, because a wrong r produces a wrong coefficient even when the arithmetic is perfect.
这类题目的步骤始终一致：写出通项，令 x 的指数等于题目要求的值，解出 r，再把 r 代回并计算乘积。不要跳过中间步骤，因为 r 取错时，即使算术完全正确，系数也是错的。
9. Counting Applications | 计数应用
rmed in C(8, 3) = 56 ways, because the order in which the names are drawn does not matter.
二项式系数可直接回答选择类问题。从 8 人的班级中选出 3 人组成委员会，共有 C(8, 3) = 56 种方式，因为名字被抽出的顺序无关紧要。
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