Unit A3:
Combinatorics单元 A3：
组合数学
The third sub-unit of Topic 1. Counting principles, permutations, combinations, and the binomial theorem — including the HL extended (rational-exponent) form. The unit IB rewards twice: once for setup, once for the right combinatorial identity.Topic 1 的第三个子单元。计数原理、排列（permutation）、组合（combination）以及二项式定理（binomial theorem）——含 HL 推广（有理指数）形式。IB 在本单元会双重给分：一次给"列式"，一次给"用对组合恒等式"。
How to use this guide本指南使用说明
Read only the dashed-gold "Cram-Mode Cheat" box at the top of each section, plus the formula boxes. The single sentence to leave with: multiply for sequential choices, $\binom{n}{r}$ for unordered, $T_{r+1} = \binom{n}{r}a^{n-r}b^r$ for binomial. Skim one worked example per section. Take the practice quiz.
只看每节顶端的金色虚线 "Cram-Mode Cheat" 速记框和公式框。一句话带走：顺序相乘、无序用 $\binom{n}{r}$、二项式用 $T_{r+1} = \binom{n}{r}a^{n-r}b^r$。每节扫一道 worked example（例题），最后做一次练习测验。
Open every ▸ Going deeper. Combinatorics is the topic where the right combinatorial argument often beats the algebraic one — especially for Pascal's identity and the binomial theorem itself. Owning both views is what separates a 5 from a 7.
打开每个 ▸ Going deeper。组合数学是"组合论证（combinatorial argument）胜过代数推导"的典型话题——尤其是帕斯卡恒等式（Pascal's identity）和二项式定理本身。同时掌握两种视角，正是 5 分到 7 分的分水岭。
Counting Principles计数原理 AHL 1.10
Addition rule: if cases $A$ and $B$ are mutually exclusive, then $|A \cup B| = |A| + |B|$. Use when cases are alternatives.
multiplication rule）：若步骤 $A$ 有 $m$ 种结果，步骤 $B$ 有 $n$ 种结果，两步连做共 $m \cdot n$ 种。适用于顺序进行的步骤。加法原理（
addition rule）：若情形 $A$ 与 $B$ 互斥（mutually exclusive），则 $|A \cup B| = |A| + |B|$。适用于互不相交的可选情形。
inclusion-exclusion）：$|A \cup B| = |A| + |B| - |A \cap B|$。
Worked Example — License plate (multiplication)例题——车牌（乘法）
Problem: A plate has 3 letters followed by 4 digits. Letters can repeat; digits cannot. How many plates are possible?
Letters: $26 \cdot 26 \cdot 26 = 26^3$. Digits (no repeats): $10 \cdot 9 \cdot 8 \cdot 7$. Plates are letters AND digits — multiply:
$$ 26^3 \cdot (10 \cdot 9 \cdot 8 \cdot 7) = 17\,576 \cdot 5040 = 88\,583\,040. $$题目：车牌由 3 个字母加 4 个数字组成。字母可重复，数字不可重复。共有多少种车牌？
字母部分：$26 \cdot 26 \cdot 26 = 26^3$。数字部分（不可重复）：$10 \cdot 9 \cdot 8 \cdot 7$。字母 AND 数字——相乘：
$$ 26^3 \cdot (10 \cdot 9 \cdot 8 \cdot 7) = 17\,576 \cdot 5040 = 88\,583\,040. $$Worked Example — Casework (addition + multiplication)例题——分类讨论（加法 + 乘法）
Problem: How many 3-digit positive integers are divisible by $5$?
Divisible by $5 \Leftrightarrow$ last digit is $0$ or $5$. These cases are disjoint, so add:
Last digit $= 0$: first digit $\in \{1, \ldots, 9\}$ (9 choices), middle digit $\in \{0, \ldots, 9\}$ (10 choices) → $9 \cdot 10 = 90$.
Last digit $= 5$: same shape → $9 \cdot 10 = 90$.
$$ \text{Total} = 90 + 90 = 180. $$题目：有多少个能被 $5$ 整除的三位正整数？
被 $5$ 整除 $\Leftrightarrow$ 末位为 $0$ 或 $5$。两情形互斥，相加即可：
末位 $= 0$：首位 $\in \{1, \ldots, 9\}$（9 选），中位 $\in \{0, \ldots, 9\}$（10 选）→ $9 \cdot 10 = 90$。
末位 $= 5$：同样形式 → $9 \cdot 10 = 90$。
$$ \text{Total} = 90 + 90 = 180. $$▸ Going deeper — Inclusion-exclusion when cases overlap▸ 深入——情形重叠时的容斥原理
When "or" cases can happen together, simple addition double-counts the overlap. The fix:
$$ |A \cup B| = |A| + |B| - |A \cap B|. $$For three sets:
$$ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|. $$Example. Out of 100 students, 40 study French, 35 Spanish, 15 study both. How many study at least one?
$$ 40 + 35 - 15 = 60. $$The IB usually phrases this as a Venn-diagram problem (see Unit D Statistics & Probability), but the underlying counting identity lives here.
当 "or" 的两种情形可以同时发生时，单纯相加会把交集多算一次。修正：
$$ |A \cup B| = |A| + |B| - |A \cap B|. $$三集合情形：
$$ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|. $$例：100 名学生中 40 人学法语，35 人学西班牙语，15 人两者都学。至少学一门的有多少人？
$$ 40 + 35 - 15 = 60. $$IB 通常把这类题包装成 Venn 图问题（见 Unit D 统计与概率），但底层的计数恒等式就是这一条。
Permutations排列 AHL 1.10
permutation）是有序的安排。从 $n$ 个互异元素中取出 $r$ 个再排序的方法数为
$$ {}^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1). $$
全取 $n$ 个排成一排：${}^nP_n = n!$。允许重复：$n^r$。含相同元素时：把 $n!$ 除以每组重复个数的阶乘。
▸ Going deeper — Why ${}^nP_r = n!/(n-r)!$▸ 深入——为何 ${}^nP_r = n!/(n-r)!$
Pick the first item: $n$ ways. The second: $n - 1$ ways (one used). The $r$-th: $n - r + 1$ ways. By the multiplication rule:
$$ {}^nP_r = n \cdot (n-1) \cdot (n-2) \cdots (n-r+1). $$Multiply numerator and denominator by $(n-r)!$:
$$ {}^nP_r = \frac{n(n-1)\cdots(n-r+1)\cdot(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!}. \;\;\blacksquare $$第一个位置：$n$ 种选法。第二个：$n - 1$ 种（已用掉一个）。第 $r$ 个：$n - r + 1$ 种。由乘法原理：
$$ {}^nP_r = n \cdot (n-1) \cdot (n-2) \cdots (n-r+1). $$分子分母同乘 $(n-r)!$：
$$ {}^nP_r = \frac{n(n-1)\cdots(n-r+1)\cdot(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!}. \;\;\blacksquare $$Forbidden positions. "$X$ not at position 1": total $n!$ minus arrangements with $X$ at position 1 ($(n-1)!$). So $n! - (n-1)! = (n-1)\cdot(n-1)!$.
Circular arrangements. $n$ distinct people around a round table: $(n-1)!$ arrangements (fix one person to remove rotational symmetry). If reflections are also identical, divide by $2$.
禁位。"$X$ 不在第 1 位"：总数 $n!$ 减去 $X$ 在第 1 位的排数 $(n-1)!$，即 $n! - (n-1)! = (n-1)\cdot(n-1)!$。
圆排列（
circular arrangement）。$n$ 个互异元素围圆桌：$(n-1)!$ 种（固定一人以消除旋转对称）。若同时认为镜像相同，再除以 $2$。
Worked Example — Arranging with constraints例题——带约束的排列
Problem: In how many ways can the letters of MATHS be arranged so that M and A are adjacent?
Glue MA into a single block. Now arrange 4 items (block, T, H, S): $4! = 24$. The block has 2 internal orders (MA, AM): multiply by $2$.
$$ 4! \cdot 2! = 24 \cdot 2 = 48. $$题目：把 MATHS 的字母重新排列，要求 M 与 A 相邻，共有多少种？
把 MA 粘成一块。现在排 4 个对象（块、T、H、S）：$4! = 24$ 种。块内有 MA、AM 两种顺序：再乘 $2$。
$$ 4! \cdot 2! = 24 \cdot 2 = 48. $$Worked Example — Identical letters例题——重复字母
Problem: How many distinguishable arrangements of the letters in MISSISSIPPI?
$11$ letters: $1$ M, $4$ I, $4$ S, $2$ P. Use the multiset formula:
$$ \frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39\,916\,800}{1 \cdot 24 \cdot 24 \cdot 2} = \frac{39\,916\,800}{1152} = 34\,650. $$题目：MISSISSIPPI 各字母排列共有多少种可区分的方式？
$11$ 个字母：$1$ 个 M，$4$ 个 I，$4$ 个 S，$2$ 个 P。用多重集排列公式：
$$ \frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39\,916\,800}{1 \cdot 24 \cdot 24 \cdot 2} = \frac{39\,916\,800}{1152} = 34\,650. $$Combinations组合 AHL 1.10
combination）是无序的选取。从 $n$ 个互异元素中选出 $r$ 个的方法数为
$$ {}^nC_r = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}. $$
$\binom{n}{r}$ 读作 "$n$ choose $r$"。与排列的联系：每个无序选法对应 $r!$ 种排序，故 ${}^nP_r = r! \cdot {}^nC_r$。
▸ Going deeper — Algebraic vs. combinatorial proofs▸ 深入——代数证法 vs. 组合证法
Take the symmetry identity $\binom{n}{r} = \binom{n}{n-r}$.
Algebraic: plug in:
$$ \binom{n}{n-r} = \frac{n!}{(n-r)!\,(n - (n-r))!} = \frac{n!}{(n-r)!\,r!} = \binom{n}{r}. $$Combinatorial: selecting $r$ to include is the same act as selecting the $n-r$ to leave out. Both descriptions count the same set of subsets. No algebra needed.
The combinatorial argument generalizes faster (e.g. to Pascal's identity below) and is what IB markschemes often want when a question says "explain" rather than "show."
以对称恒等式 $\binom{n}{r} = \binom{n}{n-r}$ 为例。
代数证：代入公式：
$$ \binom{n}{n-r} = \frac{n!}{(n-r)!\,(n - (n-r))!} = \frac{n!}{(n-r)!\,r!} = \binom{n}{r}. $$组合证（combinatorial proof）："选出 $r$ 个保留" 和 "选出 $n-r$ 个剔除" 描述的是同一件事，两种说法数的是同一族子集。无需代数。
组合证更容易推广（例如下面的帕斯卡恒等式），而且当 IB 题目用 "explain" 而非 "show" 提问时，markscheme 通常想要这种证法。
Worked Example — Committee with constraint例题——带条件的委员会
Problem: A committee of $4$ is chosen from $6$ boys and $5$ girls. How many committees have at least one girl?
Trick: complement. Total committees minus committees with no girls.
$$ \binom{11}{4} - \binom{6}{4} = 330 - 15 = 315. $$Direct case-split would also work (1 girl, 2 girls, 3 girls, 4 girls = $\binom{5}{1}\binom{6}{3} + \binom{5}{2}\binom{6}{2} + \binom{5}{3}\binom{6}{1} + \binom{5}{4}\binom{6}{0}$ = $100 + 150 + 60 + 5 = 315$ ✓), but is four times the work. Always check whether a complement is faster.
题目：从 $6$ 男 $5$ 女中选 $4$ 人组成委员会。至少有一名女生的方案数？
套路：补集。总数减去全无女生的方案数。
$$ \binom{11}{4} - \binom{6}{4} = 330 - 15 = 315. $$直接分类（1 女、2 女、3 女、4 女）也行：$\binom{5}{1}\binom{6}{3} + \binom{5}{2}\binom{6}{2} + \binom{5}{3}\binom{6}{1} + \binom{5}{4}\binom{6}{0} = 100 + 150 + 60 + 5 = 315$ ✓，但工作量是补集法的四倍。先想想"补集"是否更快。
Worked Example — Cards例题——扑克牌
Problem: A 5-card hand is dealt from a standard 52-card deck. How many hands contain exactly two aces?
Choose 2 aces from 4: $\binom{4}{2}$. Choose the remaining 3 cards from the 48 non-aces: $\binom{48}{3}$. Multiply:
$$ \binom{4}{2} \cdot \binom{48}{3} = 6 \cdot 17\,296 = 103\,776. $$题目：从标准 52 张牌中发 5 张。恰好含 2 张 A 的牌型有多少种？
从 4 张 A 中选 2 张：$\binom{4}{2}$。再从 48 张非 A 中选 3 张：$\binom{48}{3}$。相乘：
$$ \binom{4}{2} \cdot \binom{48}{3} = 6 \cdot 17\,296 = 103\,776. $$The Binomial Theorem二项式定理 SL 1.9
general term）为 $\boxed{\,T_{r+1} = \binom{n}{r} a^{n-r} b^{r}\,}$。求某变量某次方的系数：写出 $T_{r+1}$，让变量的指数等于目标值，解出 $r$，再代回。不要整体展开。
Indexed so $T_1 = a^n$ (no $b$), $T_2 = \binom{n}{1} a^{n-1}b$, etc. The subscript matches the power of $b$ plus one.下标约定：$T_1 = a^n$（不含 $b$），$T_2 = \binom{n}{1} a^{n-1}b$，依次类推。下标等于 $b$ 的次幂加 1。
▸ Going deeper — Combinatorial proof▸ 深入——组合证法
$(a+b)^n$ is the product of $n$ identical factors $(a+b)$. To distribute, from each factor pick either $a$ or $b$. A term with $b^r$ requires picking $b$ from exactly $r$ of the $n$ factors — there are $\binom{n}{r}$ ways to choose which $r$, and each contributes $a^{n-r} b^r$. Sum over $r = 0, \ldots, n$:
$$ (a+b)^n = \sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, b^{r}. \;\;\blacksquare $$This proof explains the formula rather than merely deriving it. It also generalizes: the multinomial theorem follows by counting ways to split $n$ factors into $k$ groups.
$(a+b)^n$ 是 $n$ 个相同因子 $(a+b)$ 的乘积。展开时，从每个因子中选 $a$ 或选 $b$。含 $b^r$ 的项要求恰好在 $n$ 个因子中选了 $r$ 个 $b$——选法数为 $\binom{n}{r}$，每个选法贡献 $a^{n-r} b^r$。对 $r = 0, \ldots, n$ 求和：
$$ (a+b)^n = \sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, b^{r}. \;\;\blacksquare $$这个证法解释了公式，不只是推导。它还能推广：把 $n$ 个因子分成 $k$ 组的方法数，就给出了多项式定理（multinomial theorem）。
Worked Example — Coefficient of a specific power例题——求指定次幂的系数
Problem: Find the coefficient of $x^6$ in the expansion of $\bigl(2x^2 - \tfrac{1}{x}\bigr)^9$.
Write the general term. Here $a = 2x^2$, $b = -1/x$, $n = 9$:
$$ T_{r+1} = \binom{9}{r}(2x^2)^{9-r}\!\left(-\tfrac{1}{x}\right)^{\!r} = \binom{9}{r}\, 2^{9-r}\,(-1)^r\, x^{2(9-r) - r} = \binom{9}{r}\, 2^{9-r}\,(-1)^r\, x^{18 - 3r}. $$Set exponent $= 6$. $18 - 3r = 6 \Rightarrow r = 4$.
$$ \text{Coefficient} = \binom{9}{4}\, 2^{5}\,(-1)^4 = 126 \cdot 32 \cdot 1 = 4032. $$题目：求 $\bigl(2x^2 - \tfrac{1}{x}\bigr)^9$ 展开式中 $x^6$ 的系数。
写出通项。此处 $a = 2x^2$，$b = -1/x$，$n = 9$：
$$ T_{r+1} = \binom{9}{r}(2x^2)^{9-r}\!\left(-\tfrac{1}{x}\right)^{\!r} = \binom{9}{r}\, 2^{9-r}\,(-1)^r\, x^{2(9-r) - r} = \binom{9}{r}\, 2^{9-r}\,(-1)^r\, x^{18 - 3r}. $$令指数 $= 6$：$18 - 3r = 6 \Rightarrow r = 4$。
$$ \text{Coefficient} = \binom{9}{4}\, 2^{5}\,(-1)^4 = 126 \cdot 32 \cdot 1 = 4032. $$Worked Example — Constant term例题——常数项
Problem: Find the constant term in the expansion of $\bigl(x + \tfrac{2}{x^2}\bigr)^{12}$.
$T_{r+1} = \binom{12}{r}\, x^{12 - r}\, (2/x^2)^r = \binom{12}{r}\, 2^r\, x^{12 - r - 2r} = \binom{12}{r}\, 2^r\, x^{12 - 3r}$.
Constant term has exponent $0$: $12 - 3r = 0 \Rightarrow r = 4$.
$$ T_5 = \binom{12}{4}\, 2^4 = 495 \cdot 16 = 7920. $$题目：求 $\bigl(x + \tfrac{2}{x^2}\bigr)^{12}$ 展开式中的常数项。
$T_{r+1} = \binom{12}{r}\, x^{12 - r}\, (2/x^2)^r = \binom{12}{r}\, 2^r\, x^{12 - r - 2r} = \binom{12}{r}\, 2^r\, x^{12 - 3r}$。
常数项指数为 $0$：$12 - 3r = 0 \Rightarrow r = 4$。
$$ T_5 = \binom{12}{4}\, 2^4 = 495 \cdot 16 = 7920. $$Pascal's Triangle & Identities帕斯卡三角与恒等式 SL 1.9
Pascal's triangle）每一行就是一行二项式系数 $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$。每个数等于它"肩上"两数之和（帕斯卡恒等式，Pascal's identity）。行内之和 $= 2^n$。对称：$\binom{n}{r} = \binom{n}{n-r}$。$n$ 较小时直接从三角形读取比算阶乘更快。
▸ Going deeper — Combinatorial proof of Pascal's identity▸ 深入——帕斯卡恒等式的组合证
Claim: $\displaystyle \binom{n-1}{r-1} + \binom{n-1}{r} = \binom{n}{r}$.
Setup. Count subsets of size $r$ from $\{1, 2, \ldots, n\}$. There are $\binom{n}{r}$ of them.
Split by whether they contain element $n$.
- Contains $n$: we must pick the other $r - 1$ from the $n - 1$ remaining elements — $\binom{n-1}{r-1}$ ways.
- Does not contain $n$: pick all $r$ from the $n - 1$ remaining elements — $\binom{n-1}{r}$ ways.
Every $r$-subset falls into exactly one case, so
$$ \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}. \;\;\blacksquare $$This is the prototypical "combinatorial proof": case-splitting by a chosen element, then matching counts on both sides.
命题：$\displaystyle \binom{n-1}{r-1} + \binom{n-1}{r} = \binom{n}{r}$。
构思：数 $\{1, 2, \ldots, n\}$ 中大小为 $r$ 的子集，共 $\binom{n}{r}$ 个。
按"是否含元素 $n$"分类：
- 含 $n$：剩下的 $r - 1$ 个要从其余 $n - 1$ 个元素中选——$\binom{n-1}{r-1}$ 种。
- 不含 $n$：$r$ 个全部从其余 $n - 1$ 个元素中选——$\binom{n-1}{r}$ 种。
每个 $r$ 子集恰好落在一种情形里，故
$$ \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}. \;\;\blacksquare $$这是最典型的"组合证法"：按某个特定元素分类，再两边数同一族对象。
▸ Going deeper — Why the row sum is $2^n$▸ 深入——为何行内之和为 $2^n$
Set $a = b = 1$ in the binomial theorem:
$$ (1 + 1)^n = \sum_{r=0}^{n} \binom{n}{r}\, 1^{n-r}\, 1^{r} = \sum_{r=0}^{n} \binom{n}{r}. $$The LHS is $2^n$, so the row sums to $2^n$.
Combinatorial reading: $\binom{n}{r}$ counts subsets of size $r$ from an $n$-element set; summing over all $r$ counts all subsets. There are $2^n$ subsets of an $n$-element set (each element is in or out — multiplication rule). Same number, two viewpoints.
在二项式定理中取 $a = b = 1$：
$$ (1 + 1)^n = \sum_{r=0}^{n} \binom{n}{r}\, 1^{n-r}\, 1^{r} = \sum_{r=0}^{n} \binom{n}{r}. $$左边等于 $2^n$，故该行之和为 $2^n$。
组合视角：$\binom{n}{r}$ 数的是 $n$ 元集中大小为 $r$ 的子集；对所有 $r$ 求和数的就是全部子集。$n$ 元集共有 $2^n$ 个子集（每个元素"选 / 不选"——乘法原理）。同一数，两种视角。
Extended Binomial Theorem推广二项式定理 HL AHL 1.10
radius of convergence）——是给分点。
where the generalized binomial coefficient is其中推广二项式系数为
$$ \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}. $$$(1 - x)^n$. Apply with $-x$ in place of $x$; the series alternates if you don't, so signs come from the $(-x)^r = (-1)^r x^r$ factor.
$(1 - x)^n$。把公式里的 $x$ 替换成 $-x$；不这样做级数会交错，符号由 $(-x)^r = (-1)^r x^r$ 给出。
▸ Going deeper — Why the series only converges for $|x| < 1$▸ 深入——为何级数仅在 $|x| < 1$ 内收敛
For positive-integer $n$, the binomial expansion is a finite sum (terminates at $r = n$ since $\binom{n}{r} = 0$ for $r > n$), so it converges for all $x$.
For non-integer $n$, the coefficients $\binom{n}{r} = \tfrac{n(n-1)\cdots(n-r+1)}{r!}$ never vanish, so the series is genuinely infinite. Apply the ratio test:
$$ \left|\frac{T_{r+2}}{T_{r+1}}\right| = \left|\frac{(n-r)\, x}{r+1}\right| \xrightarrow{r \to \infty} |x|. $$The series converges absolutely when $|x| < 1$ and diverges when $|x| > 1$. The boundary case $|x| = 1$ depends on $n$ — IB doesn't pursue boundary subtleties; just write "$|x| < 1$" and you're safe.
当 $n$ 为正整数时，二项展开是有限和（$r > n$ 时 $\binom{n}{r} = 0$，于 $r = n$ 处自然截断），对一切 $x$ 都收敛。
当 $n$ 非正整数时，$\binom{n}{r} = \tfrac{n(n-1)\cdots(n-r+1)}{r!}$ 永远不为零，级数真正是无穷的。用比值判别法（ratio test）：
故 $|x| < 1$ 时绝对收敛，$|x| > 1$ 时发散。边界 $|x| = 1$ 的情形取决于 $n$——IB 不深究边界细节，写 "$|x| < 1$" 即可拿分。
Worked Example — Square root via series例题——级数法求平方根
Problem: Expand $\sqrt{1 + x}$ up to and including the $x^3$ term.
$\sqrt{1+x} = (1+x)^{1/2}$. Apply the formula with $n = \tfrac{1}{2}$:
$$ (1+x)^{1/2} = 1 + \tfrac{1}{2}x + \tfrac{(1/2)(-1/2)}{2!}\,x^2 + \tfrac{(1/2)(-1/2)(-3/2)}{3!}\,x^3 + \cdots $$ $$ = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \tfrac{1}{16}x^3 - \cdots \qquad (|x| < 1). $$题目：将 $\sqrt{1 + x}$ 展开到 $x^3$ 项（含）。
$\sqrt{1+x} = (1+x)^{1/2}$。取 $n = \tfrac{1}{2}$ 代入公式：
$$ (1+x)^{1/2} = 1 + \tfrac{1}{2}x + \tfrac{(1/2)(-1/2)}{2!}\,x^2 + \tfrac{(1/2)(-1/2)(-3/2)}{3!}\,x^3 + \cdots $$ $$ = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \tfrac{1}{16}x^3 - \cdots \qquad (|x| < 1). $$Worked Example — Factor first例题——先提因式
Problem: Expand $(4 + 2x)^{-1/2}$ up to $x^2$, stating the radius of convergence.
Factor out $4^{-1/2} = \tfrac{1}{2}$:
$$ (4 + 2x)^{-1/2} = 4^{-1/2}\!\left(1 + \tfrac{x}{2}\right)^{-1/2} = \tfrac{1}{2}\!\left(1 + \tfrac{x}{2}\right)^{-1/2}. $$Apply the series with $n = -\tfrac{1}{2}$ and replace $x$ by $\tfrac{x}{2}$:
$$ \left(1 + \tfrac{x}{2}\right)^{-1/2} = 1 + (-\tfrac{1}{2})\!\cdot\!\tfrac{x}{2} + \tfrac{(-1/2)(-3/2)}{2!}\!\cdot\!\tfrac{x^2}{4} + \cdots = 1 - \tfrac{x}{4} + \tfrac{3 x^2}{32} - \cdots $$Multiply by $\tfrac{1}{2}$:
$$ (4 + 2x)^{-1/2} = \tfrac{1}{2} - \tfrac{x}{8} + \tfrac{3 x^2}{64} + \cdots $$Convergence: $\bigl|\tfrac{x}{2}\bigr| < 1 \Leftrightarrow |x| < 2$.
题目：将 $(4 + 2x)^{-1/2}$ 展开到 $x^2$ 项，并写出收敛区间。
先提 $4^{-1/2} = \tfrac{1}{2}$：
$$ (4 + 2x)^{-1/2} = 4^{-1/2}\!\left(1 + \tfrac{x}{2}\right)^{-1/2} = \tfrac{1}{2}\!\left(1 + \tfrac{x}{2}\right)^{-1/2}. $$取 $n = -\tfrac{1}{2}$，并把公式里的 $x$ 替换为 $\tfrac{x}{2}$：
$$ \left(1 + \tfrac{x}{2}\right)^{-1/2} = 1 + (-\tfrac{1}{2})\!\cdot\!\tfrac{x}{2} + \tfrac{(-1/2)(-3/2)}{2!}\!\cdot\!\tfrac{x^2}{4} + \cdots = 1 - \tfrac{x}{4} + \tfrac{3 x^2}{32} - \cdots $$再乘上 $\tfrac{1}{2}$：
$$ (4 + 2x)^{-1/2} = \tfrac{1}{2} - \tfrac{x}{8} + \tfrac{3 x^2}{64} + \cdots $$收敛：$\bigl|\tfrac{x}{2}\bigr| < 1 \Leftrightarrow |x| < 2$。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- ${}^nP_r = \tfrac{n!}{(n-r)!}$ and $\binom{n}{r} = \tfrac{n!}{r!(n-r)!}$
- Pascal's identity and the row-sum identity
- $T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^{r}$ for binomial
- Extended binomial — formula and the $|x| < 1$ condition HL
- Pascal rows up to $n = 6$ at sight
- Multiset permutation $\tfrac{n!}{n_1!\,n_2!\cdots n_k!}$
- ${}^nP_r = \tfrac{n!}{(n-r)!}$ 与 $\binom{n}{r} = \tfrac{n!}{r!(n-r)!}$
- 帕斯卡恒等式以及行和恒等式
- 二项式通项 $T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^{r}$
- 推广二项式——公式及收敛条件 $|x| < 1$ HL
- 帕斯卡三角至 $n = 6$ 行能默写
- 多重集排列公式 $\tfrac{n!}{n_1!\,n_2!\cdots n_k!}$
- When to multiply (sequential) vs. add (mutually exclusive)
- Why ${}^nP_r = r!\cdot\binom{n}{r}$ — orderings of an unordered selection
- Pascal's identity via case-split on a chosen element
- Combinatorial proof of the binomial theorem (pick $b$ from $r$ factors)
- Why the extended binomial fails for $|x| \ge 1$ (ratio test) HL
- Complement-counting: when is $|U| - |\text{not-}A|$ faster than direct?
- 何时用乘法（顺序）何时用加法（互斥）
- 为何 ${}^nP_r = r!\cdot\binom{n}{r}$——把"无序选取"再排序
- 帕斯卡恒等式：按某元素分类
- 二项式定理的组合证（从 $r$ 个因子中选 $b$）
- 推广二项式为何在 $|x| \ge 1$ 时失效（比值判别） HL
- 补集计数：何时 $|U| - |\text{非 } A|$ 比直接分类更快
Common Pitfalls常见陷阱
2. Forgetting the sign on $(-x)^r$ — the $(-1)^r$ factor changes parity.
3. Treating arrangements as combinations (or vice versa). Read for "order" cues.
4. Using addition rule on overlapping cases without inclusion-exclusion.
5. Failing to factor before applying the extended binomial — $(2 + x)^n \ne$ direct $(1 + \text{stuff})^n$ form.
6. Omitting the radius of convergence on Paper 1 extended-binomial questions.
7. Multiset permutations: forgetting to divide by all repeat-factorials, e.g. dividing only once for MISSISSIPPI.
8. Off-by-one on circular arrangements: $n$ around a circle is $(n-1)!$, not $n!$.
2. 忘记 $(-x)^r$ 的符号——$(-1)^r$ 会改变奇偶号。
3. 把排列当组合（或反之）。注意题面里的"顺序"提示词。
4. 在有重叠的情形上直接用加法，忘了容斥原理。
5. 不先提因式就用推广二项式——$(2 + x)^n$ 不是 $(1 + \cdots)^n$ 的直接形式。
6. Paper 1 推广二项式题忘写收敛区间。
7. 多重集排列：忘记把所有重复个数的阶乘都除掉，比如 MISSISSIPPI 只除一个。
8. 圆排列差一：$n$ 人围成圆是 $(n-1)!$，不是 $n!$。
Paper 2 (calc): Larger committee/probability counts (you'll evaluate factorials on calc); approximations using extended binomial (e.g. $\sqrt{1.04}$ via the series with $x = 0.04$).
Paper 2（可用计算器）：较大的委员会 / 概率计数（用计算器算阶乘）；用推广二项式做近似（如 $\sqrt{1.04}$ 取 $x = 0.04$）。
Flashcards闪卡
Mutually exclusive OR → add顺序 AND → 乘
互斥 OR → 加
$|x| < 1$
Unit A3 — Practice Quiz单元 A3——练习测验
Ten mixed-difficulty items. Your score updates in real time at the top of the page. Aim for 8/10 before exam day.十道难度不一的题。得分实时显示在页面顶部。考前目标 8/10。
Readiness Checklist备考清单
Click each item you've mastered. Aim for 100% before exam day. Items marked HL are HL-only.点击你已经掌握的条目。考前目标 100%。标有 HL 的为 HL 专属内容。
- Use multiplication and addition rules; recognize when cases overlap熟练使用乘法与加法原理；识别情形是否重叠
- State and apply ${}^nP_r = n!/(n-r)!$, including small-$r$ shortcuts写出并应用 ${}^nP_r = n!/(n-r)!$，含小 $r$ 时的快速算法
- Handle multiset permutations and circular arrangements能处理多重集排列与圆排列
- State and apply $\binom{n}{r}$; use the symmetry $\binom{n}{r} = \binom{n}{n-r}$写出并应用 $\binom{n}{r}$；活用对称性 $\binom{n}{r} = \binom{n}{n-r}$
- Use complement counting when faster than direct casework能在补集法比直接分类更快时果断切换
- Apply the binomial theorem to find specific terms or coefficients用二项式定理求指定项或指定系数
- Distinguish "$r$-th term" from "$T_r$" — index correctly区分"第 $r$ 项"与 "$T_r$"——下标对齐
- Reproduce the combinatorial proof of the binomial theorem能复现二项式定理的组合证
- Recognize Pascal rows up to $n = 6$ and use Pascal's identity识别帕斯卡三角前 $n = 6$ 行并运用帕斯卡恒等式
- Prove the row sum is $2^n$ by both algebraic and combinatorial routes能用代数法与组合法分别证明行和为 $2^n$
- Reproduce the combinatorial proof of Pascal's identity (case-split on element $n$)能复现帕斯卡恒等式的组合证（按元素 $n$ 分类）
- HL Apply the extended binomial theorem to rational/negative $n$能对有理 / 负指数 $n$ 应用推广二项式定理
- HL Factor $(a + bx)^n$ before applying the series; state radius of convergence展开 $(a + bx)^n$ 前先提因式；写出收敛区间
- HL Use the extended binomial to approximate roots and reciprocals用推广二项式做开方与倒数的近似
IB Paper-Style PracticeIB 试卷风格练习
IB exam-style questions across Paper 1A (short response, no calc), Paper 1B (extended response, no calc), Paper 2 (calculator), and a Paper 3 HL extended exploration. EMH difficulty mix. Mark-by-mark solutions (M1 / A1 / R1) live in the separate solutions file. Use this after the in-page quiz and flashcards.
IB 考试风格题，涵盖 Paper 1A（短答，无计算器）、Paper 1B（长答，无计算器）、Paper 2（可用计算器），并附 Paper 3 HL 长题探究。难度按 EMH 分级。逐分（M1 / A1 / R1）解答见独立的解答文档。建议在做完本页测验与闪卡后再来。