PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 28 marks不可使用计算器 · 简答题 · 28 分
Section A — Short ResponseA 节 —— 简答题
Show all working in the space below each question. Method marks dominate — partial credit is generous when the set-up is clean. No calculator permitted. Keep expressions exact; do not convert factorials or binomial coefficients to decimals.在每题下方的空白处写出全部解题过程。方法分（method marks）权重最大 —— 只要起手式（set-up）干净，过程分给得较为宽松。不可使用计算器。表达式保留精确形式；不要将阶乘（factorial）或二项式系数（binomial coefficient）转为小数。
Q1MEDIUMPaper 1A1.9 Binomial — Term Hunt[6 marks]
Consider the expansion of $\displaystyle\left(2x^{2} - \dfrac{1}{x}\right)^{9}$ in descending powers of $x$.考虑 $\displaystyle\left(2x^{2} - \dfrac{1}{x}\right)^{9}$ 按 $x$ 的降幂展开式（binomial expansion）。
(a)Write down the general term $T_{r+1}$ in this expansion. Simplify so it is a single power of $x$ multiplied by a numerical coefficient.写出展开式的通项（general term）$T_{r+1}$。化简为 $x$ 的单一幂次乘以一个数值系数（coefficient）。[2]
(b)Hence find the coefficient of $x^{6}$.由此求 $x^{6}$ 的系数。[2]
(c)Find the term in the expansion that is independent of $x$.求展开式中不含 $x$ 的项（term independent of $x$）。[2]
Q2MEDIUMPaper 1A1.10 Multiset Perms + Gap Method[7 marks]
Consider the letters of the word MISSISSIPPI.考虑单词 MISSISSIPPI 中的字母。
(a)Show that the number of distinguishable arrangements of all eleven letters is $\dfrac{11!}{4!\,4!\,2!}$, and evaluate this number.证明：所有 11 个字母的可区分排列（distinguishable arrangements）数为 $\dfrac{11!}{4!\,4!\,2!}$，并算出该值。[2]
(b)Find the number of distinguishable arrangements in which no two I's are adjacent. Justify your method clearly.求没有任何两个 I 相邻的可区分排列数。清晰说明所用方法。[5]
Three consecutive binomial coefficients $\dbinom{n}{r}$, $\dbinom{n}{r+1}$, $\dbinom{n}{r+2}$ are in the ratio三个相邻的二项式系数 $\dbinom{n}{r}$、$\dbinom{n}{r+1}$、$\dbinom{n}{r+2}$ 满足如下比例（ratio）：
(a)By differentiating both sides with respect to $x$ and substituting a suitable value of $x$, prove that对两边关于 $x$ 求导，并代入合适的 $x$ 值，证明[3]
$$ \sum_{r=1}^{n} r\,\binom{n}{r} \;=\; n\,2^{\,n-1}. $$
(b)By differentiating twice and substituting a suitable value of $x$, prove that对两边求两次导数，并代入合适的 $x$ 值，证明[3]
$$ \sum_{r=2}^{n} r(r-1)\,\binom{n}{r} \;=\; n(n-1)\,2^{\,n-2}. $$
(c)Hence, or otherwise, show that for $n \ge 2$由此（或采用其他方法）证明：当 $n \ge 2$ 时[2]
$$ \sum_{r=0}^{n} r^{2}\,\binom{n}{r} \;=\; n(n+1)\,2^{\,n-2}. $$
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 36 marks不可使用计算器 · 长答题 · 36 分
Section B — Extended ResponseB 节 —— 长答题
Take time to set up cleanly — structure carries method marks. State the binomial formulas you cite. Carry expressions in exact form; do not approximate. Justify each combinatorial argument with a short sentence describing what is being counted.认真梳理思路再下笔 —— 结构清晰本身就承担方法分。引用的二项公式要明确写出。表达式保留精确形式，不要做近似。每一个计数论证（combinatorial argument）都用一句话说明"在数什么"。
Q5HARDPaper 1B1.9 Binomial — System & Largest Coefficient[12 marks]
Consider the expansion of $(1 + ax)^{n}$ in ascending powers of $x$, where $a$ is a non-zero real constant and $n$ is a positive integer with $n \ge 3$.考虑 $(1 + ax)^{n}$ 按 $x$ 的升幂展开，其中 $a$ 为非零实数常数，$n$ 为正整数且 $n \ge 3$。
(a)Write down the general term $T_{r+1}$ in terms of $n$, $a$, $r$.用 $n$、$a$、$r$ 写出通项 $T_{r+1}$。[1]
(b)Show that the coefficient of $x^{k}$ in this expansion is $\dbinom{n}{k}\,a^{k}$.证明该展开式中 $x^{k}$ 的系数为 $\dbinom{n}{k}\,a^{k}$。[2]
(c)The coefficient of $x$ in this expansion is $24$ and the coefficient of $x^{2}$ is $252$.该展开式中 $x$ 的系数为 $24$，$x^{2}$ 的系数为 $252$。
(i)Use the result of (b) to write down two equations in $n$ and $a$.利用 (b) 的结论，写出关于 $n$ 与 $a$ 的两个方程。[2]
(ii)Show that $n = 8$ and find the value of $a$.证明 $n = 8$，并求 $a$ 的值。[3]
(d)Hence find the coefficient of $x^{3}$.由此求 $x^{3}$ 的系数。[1]
(e)Using the ratio $T_{r+2}/T_{r+1}$, determine the value of $r$ for which the term $T_{r+1}$ in the expansion of $(1+ax)^{8}$ (with the value of $a$ from (c)) is largest, and state the largest coefficient.利用比值 $T_{r+2}/T_{r+1}$，确定 $(1+ax)^{8}$ 展开式（$a$ 取 (c) 中求得之值）中使 $T_{r+1}$ 最大的 $r$ 值，并写出最大系数。[3]
(a)Verify the identity for $r = 2$ and $n = 5$ by evaluating both sides as integers.取 $r = 2$、$n = 5$，将两边作为整数分别求值，验证该恒等式。[3]
(b)Prove the identity by counting the $(r+1)$-element subsets of $\{1, 2, 3, \ldots, n+1\}$ in two different ways. Begin by classifying each such subset by its largest element.通过对集合 $\{1, 2, 3, \ldots, n+1\}$ 的所有 $(r+1)$-元子集（subset）用两种不同方法计数，证明该恒等式。先按每个子集的最大元素进行分类。[5]
(c)Use the identity to evaluate $\displaystyle\sum_{k=3}^{20}\binom{k}{3}$, expressing your answer as a single binomial coefficient and as an integer.利用该恒等式计算 $\displaystyle\sum_{k=3}^{20}\binom{k}{3}$，答案分别写成单个二项式系数与整数形式。[2]
(d)Noting that $k = \dbinom{k}{1}$, use the identity to prove that注意到 $k = \dbinom{k}{1}$，利用该恒等式证明[3]
$$ 1 + 2 + 3 + \cdots + n \;=\; \dfrac{n(n+1)}{2}. $$
Q11HARDPaper 1B1.10 Lattice Paths — Combinations on a Grid[11 marks]
Consider the lattice paths from the point $(0,0)$ to the point $(5,4)$ in the Cartesian plane that are monotone: at each step a path moves one unit right ($R$: $(x, y)\to(x+1, y)$) or one unit up ($U$: $(x, y)\to(x, y+1)$). Each such path is uniquely encoded by the sequence of $R$'s and $U$'s used.考虑直角坐标平面中从点 $(0,0)$ 到点 $(5,4)$ 的单调格点路径（monotone lattice paths）：每一步要么向右移动一格（$R$：$(x, y)\to(x+1, y)$），要么向上移动一格（$U$：$(x, y)\to(x, y+1)$）。每条路径由所使用的 $R$ 与 $U$ 序列唯一表示。
(a)Explain why the total number of such monotone paths is $\dbinom{9}{4}$, and evaluate it.说明为何这种单调路径的总数为 $\dbinom{9}{4}$，并算出该值。[2]
(b)Find the number of monotone paths from $(0,0)$ to $(5,4)$ that pass through the point $(3,2)$.求从 $(0,0)$ 到 $(5,4)$ 且经过点 $(3,2)$ 的单调路径数。[3]
(c)Hence find the number of monotone paths that do not pass through $(3,2)$.由此求不经过 $(3,2)$ 的单调路径数。[1]
(d)Find the number of monotone paths that pass through $(3,2)$ but do not pass through $(1,1)$.求经过 $(3,2)$ 但不经过 $(1,1)$ 的单调路径数。[4]
(e)One monotone path is chosen uniformly at random from all the paths in part (a). State, with brief justification, the probability that this path passes through both $(1,1)$ and $(3,2)$.从 (a) 中所有单调路径中均匀随机选取一条。简要说明该路径同时经过 $(1,1)$ 与 $(3,2)$ 的概率。[1]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 36 marks可使用计算器 · 混合题型 · 36 分
Paper 2 — Calculator Permitted第二卷 —— 允许使用计算器
A graphing calculator is required. Give exact answers (integers or fractions) where reasonable; otherwise correct to $3$ significant figures unless specified. Show set-up steps even when the final evaluation is done on the GDC — method marks are still on the line.需要图形计算器（GDC）。能给出精确答案（整数或分数）的题目就给精确答案；其余除特别说明外保留 $3$ 位有效数字。即便最终求值在 GDC 上完成，也要写出列式步骤 —— 方法分依旧计入。
A committee of $4$ is selected at random from a group of $12$ students, consisting of $5$ IB Diploma students and $7$ AP students. Each of the $\dbinom{12}{4}$ possible committees is equally likely.从 $12$ 名学生（$5$ 名 IB 学生与 $7$ 名 AP 学生）中随机选出一个 $4$ 人委员会。共 $\dbinom{12}{4}$ 种可能的委员会，每种等可能。
(a)Find the probability that the committee contains exactly $2$ IB students. Give your answer as an exact fraction in lowest terms.求委员会恰好有 $2$ 名 IB 学生的概率。答案以最简精确分数形式给出。[2]
(b)Find the probability that the committee contains at least $2$ IB students.求委员会至少有 $2$ 名 IB 学生的概率。[3]
(c)Given that the committee contains at least one IB student, find the probability that it contains exactly $2$ IB students.在委员会至少有一名 IB 学生的条件下，求其恰好有 $2$ 名 IB 学生的概率（条件概率，conditional probability）。[2]
Q8MEDIUMPaper 21.10 Linear Perms — Gap Method[7 marks]
A class of $6$ boys and $4$ girls is to be seated in a row of $10$ chairs. Treat all students as distinguishable.$6$ 名男生与 $4$ 名女生组成的班级要在 $10$ 张椅子排成的一行中就座。所有学生视为可区分的。
(a)Find the total number of seating arrangements.求座位安排的总数。[1]
(b)Find the number of arrangements in which no two girls sit next to each other.求任意两名女生都不相邻的安排数（插空法，gap method）。[3]
(c)Hence find the probability that, in a random arrangement, no two girls are adjacent.由此求随机安排下任意两名女生都不相邻的概率。[1]
(d)Find the number of arrangements in which a particular girl $G$ sits between two particular boys $B_{1}$ and $B_{2}$ (so the three of them form a consecutive block $B_{1} \,G\, B_{2}$ or $B_{2}\, G \, B_{1}$, somewhere in the row).求某位指定女生 $G$ 恰好坐在两位指定男生 $B_{1}$ 与 $B_{2}$ 中间的安排数（即三人构成连续块 $B_{1} \,G\, B_{2}$ 或 $B_{2}\, G \, B_{1}$，位于行中任意位置）。[2]
Ten people, including three sisters Alice, Beth, and Cara, are to be seated around a round table with $10$ chairs. Two seatings are considered identical when one can be rotated to the other.十人（其中包含三姐妹 Alice、Beth、Cara）围一张有 $10$ 张椅子的圆桌就座。两种座次若可通过旋转互相对应，则视为同一种安排（圆排列，circular permutation）。
(a)Find the total number of distinguishable seating arrangements.求可区分的座位安排总数。[1]
(b)Find the number of arrangements in which the three sisters all sit together (as a consecutive block of three).求三姐妹全部相邻（构成一个三人连续块）的安排数。[2]
(c)Find the number of arrangements in which no two of the three sisters sit next to each other.求三姐妹中任意两人均不相邻的安排数。[4]
This question uses the extended (rational-exponent) binomial theorem to approximate $\sqrt{2}$. The series in (a) is the workhorse; parts (b)–(d) apply a scaling trick because the naive substitution $x=1$ lies outside the radius of convergence.本题利用广义二项定理（extended binomial theorem，允许有理数指数）逼近 $\sqrt{2}$。(a) 中的级数是核心工具；(b)–(d) 使用一个换元技巧（scaling trick），因为直接代入 $x=1$ 落在收敛半径（radius of convergence）之外。
(a)Use the extended binomial theorem to find the first four non-zero terms of the expansion of $\sqrt{1+x}$ in ascending powers of $x$, and state the values of $x$ for which the expansion is valid.利用广义二项定理求 $\sqrt{1+x}$ 按 $x$ 升幂展开的前四个非零项，并写出展开式成立的 $x$ 取值范围。[3]
(b)Explain in one sentence why the value $x = 1$ cannot be substituted directly into your expansion to approximate $\sqrt{2}$.用一句话说明：为何不能直接将 $x = 1$ 代入展开式来逼近 $\sqrt{2}$。[1]
(c)Show that $\sqrt{2}$ can be written as $\dfrac{7}{5}\,\sqrt{1+r}$ where $r = \dfrac{1}{49}$.证明 $\sqrt{2}$ 可以写成 $\dfrac{7}{5}\,\sqrt{1+r}$，其中 $r = \dfrac{1}{49}$。[1]
(d)Hence, using the first four terms of your expansion from (a) with $x = \dfrac{1}{49}$, approximate $\sqrt{2}$ to $5$ decimal places. Compare with the GDC value of $\sqrt{2}$ to $5$ decimal places and state the magnitude of the error.由此，在 (a) 的展开式中取前四项并代入 $x = \dfrac{1}{49}$，将 $\sqrt{2}$ 估计到 $5$ 位小数。与 GDC 给出的 $5$ 位小数 $\sqrt{2}$ 值作比较，并写出误差的数量级。[2]
A permutation of the digits $\{1, 2, 3, \ldots, 10\}$ is selected uniformly at random from the $10!$ possible permutations.从数字集合 $\{1, 2, 3, \ldots, 10\}$ 的 $10!$ 种排列（permutation）中均匀随机选取一种。
(a)Find the probability that the digits $1$ and $2$ appear in adjacent positions in the permutation (in either order).求数字 $1$ 与 $2$ 在排列中处于相邻位置（顺序任意）的概率。[2]
(b)Find the probability that the digit $1$ appears somewhere to the left of the digit $2$ in the permutation (not necessarily adjacent). Briefly justify your answer using a symmetry argument.求数字 $1$ 在排列中位于数字 $2$ 左侧某位置（不必相邻）的概率。用对称性论证（symmetry argument）简要说明答案。[2]
(c)Find the probability that the digits $1$, $2$, $3$ appear in increasing positional order — that is, $1$ is somewhere to the left of $2$, which is somewhere to the left of $3$.求数字 $1$、$2$、$3$ 在排列中按位置依次递增出现的概率 —— 即 $1$ 在 $2$ 的左侧某位置，$2$ 又在 $3$ 的左侧某位置。[2]
(d)Let $k$ be a fixed integer with $2 \le k \le 10$. Find, with justification, the probability that an arbitrary $k$ specified digits from $\{1, 2, \ldots, 10\}$ appear in increasing positional order in the random permutation.设 $k$ 为固定整数，$2 \le k \le 10$。给出（并说明理由）任意指定的 $k$ 个来自 $\{1, 2, \ldots, 10\}$ 的数字在随机排列中按位置依次递增出现的概率。[2]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 16 marks可使用计算器 · HL 长题探究 · 16 分
Paper 3 — HL Extended Problem第三卷 —— HL 长题探究
A graphing calculator is required. Method marks are heavily weighted. Show full reasoning at every step; partial credit is generous.需要图形计算器（GDC）。方法分权重很高。每一步都要写出完整推理；过程分（partial credit）给得较为宽松。
This question develops Vandermonde's identity from Pascal's identity, then uses it to evaluate a famous closed-form sum of squared binomial coefficients.本题从帕斯卡恒等式（Pascal's identity）出发推出范德蒙恒等式（Vandermonde's identity），并用它求一个著名的二项式系数平方和（sum of squared binomial coefficients）的闭合形式。
(a)State Pascal's identity for $\binom{n}{r}$. Give a one-sentence committee interpretation explaining why it holds.写出 $\binom{n}{r}$ 的帕斯卡恒等式。用一句话给出"委员会选择"（committee selection）的组合解释，说明该恒等式为何成立。[2]
(b)Vandermonde's identity asserts that for non-negative integers $m, n, r$ with $r \le m + n$,范德蒙恒等式断言：对任意非负整数 $m, n, r$（$r \le m + n$）有
$$ \binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k}. $$
Verify the identity in the case $m = n = 3$, $r = 2$ by computing both sides explicitly.在 $m = n = 3$、$r = 2$ 的情形下，显式计算两边以验证该恒等式。[3]
(c)Prove Vandermonde's identity by a combinatorial argument. Hint: count $r$-element subsets of a set partitioned into two disjoint blocks of sizes $m$ and $n$, conditioning on how many elements come from the first block.用组合论证法证明范德蒙恒等式。提示：在一个被划分为大小分别为 $m$ 与 $n$ 的两个不相交块（disjoint blocks）的集合中，按照来自第一块的元素个数分类，计数所有 $r$ 元子集。[4]
(d)By specialising Vandermonde's identity to $m = n$ and $r = n$, and using the symmetry $\binom{n}{n-k} = \binom{n}{k}$, derive the closed-form identity在范德蒙恒等式中取 $m = n$、$r = n$，并利用对称性 $\binom{n}{n-k} = \binom{n}{k}$，推导闭合恒等式
$$ \sum_{k=0}^{n} \binom{n}{k}^{\!2} = \binom{2n}{n}. $$
[3]
(e)Use your GDC to compute both $\displaystyle\sum_{k=0}^{5} \binom{5}{k}^{\!2}$ and $\binom{10}{5}$ as numerical checks of the identity in (d). State both values and confirm they agree.在 GDC 上计算 $\displaystyle\sum_{k=0}^{5} \binom{5}{k}^{\!2}$ 与 $\binom{10}{5}$ 的数值，作为对 (d) 中恒等式的检验。给出两个数值并确认它们相等。[2]
(f)Without re-proving the identity, write down a closed-form expression for不必重新证明，写出下式的闭合形式表达式：
$$ \sum_{k=0}^{n-1} \binom{n}{k}\binom{n}{k+1} $$
by applying Vandermonde's identity to a suitable choice of $m$, $n$, $r$. State your $m$, $n$, $r$ explicitly and justify the symmetry step you use.方法：对范德蒙恒等式选取合适的 $m$、$n$、$r$。明确写出你的 $m$、$n$、$r$，并说明所用的对称性步骤。[2]