PART I · SHORT RESPONSE第一部分 · 简答题SAT-style MCQ + Provincial short answer · 18 marksSAT 风格选择题 + 省考简答题 · 18 分
Section A · Short ResponseA 节 · 简答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. Before any count, state whether order matters (permutation) or order does not (combination). No calculator on Q1–Q4; calculator permitted on Q5.题型混合：选择题与简答题。选择题请圈出字母，并在空白处保留可供阅卷复核的过程。任何计数之前，先表明是否计较顺序（排列）或不计较顺序（组合）。Q1–Q4 不得使用计算器；Q5 可用计算器。
A cafeteria offers $4$ sandwiches, $3$ sides, and $5$ drinks. A meal consists of one sandwich, one side, and one drink. How many distinct meals are possible?一家自助餐厅提供 $4$ 种三明治、$3$ 种配菜和 $5$ 种饮品。一份套餐由一份三明治、一份配菜与一份饮品组成。可组成多少种不同的套餐？
Which of the following is equal to $\binom{15}{4}$?下列哪一项等于 $\binom{15}{4}$？
(A) $\binom{15}{3}$
(B) $\binom{15}{11}$
(C) $\binom{4}{15}$
(D) $\binom{11}{4}$
Q4MEDIUM🇨🇦 ONON Provincial-style§3 Combinations · MDM4U Counting and Probability[4 marks]
A school's debate club has $12$ members. Three of them must be chosen to attend a tournament.某学校辩论俱乐部共有 $12$ 名成员。需从中选出 $3$ 人参加锦标赛。
(a)If the three attendees are sent as an undifferentiated team (no roles), how many ways are there to choose them? State whether you used ${}_{n}P_{r}$ or $\binom{n}{r}$ and justify in one sentence.若三位代表作为一支不区分角色的团队前往（无角色之分），有多少种选法？说明使用了 ${}_{n}P_{r}$ 还是 $\binom{n}{r}$，并用一句话解释理由。[2]
(b)If, instead, the three attendees are to be assigned the distinct roles of captain, first speaker, and second speaker, how many ways are there?若三位代表反而要被分配队长、一辩、二辩三个不同角色，有多少种安排方法？[1]
(c)Verify numerically that your answer to (b) equals $3!$ times your answer to (a), and explain the factor of $3!$ combinatorially.数值上验证 (b) 的结果等于 $3!$ 乘以 (a) 的结果，并从组合学角度解释 $3!$ 这个因子。[1]
Q5MEDIUM🇨🇦 ABAB Provincial-style§2 Identical-Elements Permutations · AB Math 30-1 SO 2 (indicator 2.6)[5 marks]
Consider the word STATISTICS ($10$ letters: $3$ S, $3$ T, $1$ A, $2$ I, $1$ C).考虑单词 STATISTICS（共 $10$ 个字母：$3$ 个 S、$3$ 个 T、$1$ 个 A、$2$ 个 I、$1$ 个 C）。
(a)State the identical-elements formula for the number of distinguishable letter arrangements of a word with repeated letters, in terms of the total letter count $n$ and the multiplicities $k_{1}, k_{2}, \ldots, k_{m}$.用总字母数 $n$ 与重复次数 $k_{1}, k_{2}, \ldots, k_{m}$，写出含重复字母单词可区分排列数的"相同元素"公式。[1]
(b)Compute the number of distinguishable letter arrangements of STATISTICS. Give the exact integer.计算 STATISTICS 的可区分字母排列总数，给出精确整数。[2]
(c)If the word were instead STATS ($5$ letters: $2$ S, $2$ T, $1$ A), how many distinguishable arrangements would there be?若单词改为 STATS（共 $5$ 个字母：$2$ 个 S、$2$ 个 T、$1$ 个 A），可区分的排列共有多少个？[2]
PART II · EXTENDED RESPONSE第二部分 · 长答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 35 分
Section B · Extended ResponseB 节 · 长答题
Show every algebraic step. State the general term $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$ before substituting in any binomial-expansion problem. No calculator on Q6–Q9 unless noted.每一步代数过程都要写出。任何二项展开题在代入数值前，先写出通项 $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$。除非特别注明，Q6–Q9 不得使用计算器。
Q6MEDIUM🇨🇦 BCBC Provincial-style§3 Committee Counts · BC PC 12 Permutations/Combinations[8 marks]
A committee of $5$ is to be chosen from a group of $8$ women and $6$ men.从 $8$ 名女性与 $6$ 名男性组成的群体中选出一个 $5$ 人委员会。
(a)How many committees are possible with no restriction on composition?如对人员构成无任何限制，共有多少种委员会？[2]
(b)How many committees contain exactly $3$ women and $2$ men?恰好含 $3$ 名女性与 $2$ 名男性的委员会有多少个？[3]
(c)How many committees contain at least $4$ women?至少含 $4$ 名女性的委员会有多少个？[3]
Q7MEDIUM🇨🇦 ABAB Provincial-style§5 Binomial Theorem · AB Math 30-1 SO 4 (indicators 4.1, 4.5)[8 marks]
Consider the expansion of $(2x + 3)^{5}$ via the Binomial Theorem.用二项式定理考察 $(2x + 3)^{5}$ 的展开。
(a)Write row $5$ of Pascal's triangle.写出帕斯卡三角形第 $5$ 行。[1]
(b)Write out the general term $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$ for $(2x + 3)^{5}$, with $a = 2x$, $b = 3$, $n = 5$, $r = 0, 1, \ldots, 5$.写出 $(2x + 3)^{5}$ 的通项 $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$，其中 $a = 2x$、$b = 3$、$n = 5$、$r = 0, 1, \ldots, 5$。[2]
(c)Compute the full expansion of $(2x + 3)^{5}$ in expanded standard form, simplifying each coefficient.写出 $(2x + 3)^{5}$ 的完整展开（标准展开形式），化简每一项系数。[4]
(d)State the sum of all coefficients in the expansion, and check it equals $(2(1) + 3)^{5} = 5^{5} = 3125$.写出展开式所有系数之和，并验证它等于 $(2(1) + 3)^{5} = 5^{5} = 3125$。[1]
Consider the expansion of $(2x - 3)^{12}$.考察 $(2x - 3)^{12}$ 的展开。
(a)State the general term $T_{r+1}$ of $(2x - 3)^{12}$ as a function of $r$, where $r \in \{0, 1, 2, \ldots, 12\}$. Simplify so that the dependence on $x$ is in a single power $x^{12-r}$.将 $(2x - 3)^{12}$ 的通项 $T_{r+1}$ 写成 $r$ 的函数，其中 $r \in \{0, 1, 2, \ldots, 12\}$。化简使 $x$ 仅以单一幂次 $x^{12-r}$ 出现。[3]
(b)Find the value of $r$ that produces the term containing $x^{7}$.求出对应含 $x^{7}$ 项的 $r$ 值。[2]
(c)Compute the coefficient of $x^{7}$ in the expansion of $(2x - 3)^{12}$. Give the exact signed integer.求 $(2x - 3)^{12}$ 展开式中 $x^{7}$ 的系数，给出带符号的精确整数。[3]
(d)State the sign of your coefficient and explain in one sentence why this sign is consistent with the pattern $T_{r+1}$ alternates sign as $r$ increases by one.写出该系数的符号，并用一句话解释为何这一符号与 $r$ 每增加 $1$ 时 $T_{r+1}$ 符号交替的规律相符。[1]
Q9HARDHonors🇺🇸 US🇨🇦 BCAP-feeder FRQ§6 Term Extraction (negative powers) · BC PC 12 / HSA-APR.C.5 (+)[10 marks]
Consider the expansion of $\left(x + \dfrac{2}{x}\right)^{10}$, valid for $x \ne 0$.考察 $\left(x + \dfrac{2}{x}\right)^{10}$ 的展开（$x \ne 0$）。
(a)State the general term $T_{r+1}$ as a function of $r$, where $r \in \{0, 1, 2, \ldots, 10\}$. Simplify so that the dependence on $x$ is in a single power $x^{10-2r}$.将通项 $T_{r+1}$ 写成 $r$ 的函数，其中 $r \in \{0, 1, 2, \ldots, 10\}$。化简使 $x$ 仅以单一幂次 $x^{10-2r}$ 出现。[2]
(b)Find the value of $r$ that produces the constant term (the term with $x^{0}$).求出对应常数项（即 $x^{0}$ 项）的 $r$ 值。[2]
(c)Compute the constant term. Give the exact integer.求出常数项，给出精确整数。[3]
(d)Find the value of $r$ that produces the term containing $x^{4}$, and compute the coefficient of $x^{4}$.求出对应含 $x^{4}$ 项的 $r$ 值，并求 $x^{4}$ 的系数。[2]
(e)Explain in one sentence why only even powers of $x$ can appear in this expansion (cite the form of the exponent of $x$ in $T_{r+1}$).用一句话解释为何此展开式中只能出现 $x$ 的偶数次幂（请引用 $T_{r+1}$ 中 $x$ 指数的形式）。[1]
Name your sample space and the choice rule (with or without replacement; ordered or unordered) before writing any count. State each combinatorial identity in symbols before applying it. Calculator permitted throughout Part III.写出任何计数前，先指明样本空间与选取规则（是否放回；是否有序）。每个组合恒等式在使用前先以符号写出。第三部分全程可使用计算器。
A province issues licence plates with $3$ uppercase letters from $\{A, B, \ldots, Z\}$ followed by $3$ digits from $\{0, 1, \ldots, 9\}$. The letters $I, O$ and the digit $0$ are excluded as confusing (so there are effectively $24$ usable letters and $9$ usable digits).某省份发放的车牌由 $\{A, B, \ldots, Z\}$ 中的 $3$ 个大写字母，后接 $\{0, 1, \ldots, 9\}$ 中的 $3$ 个数字组成。字母 $I, O$ 与数字 $0$ 因易混淆而被排除（故实际可用字母 $24$ 个，可用数字 $9$ 个）。
(a)How many distinct plates are possible if repetition is allowed throughout?若允许全程重复，可生成多少种不同车牌？[2]
(b)How many plates are possible if no letter and no digit may be repeated?若字母不得重复且数字也不得重复，可生成多少种车牌？[2]
(c)A driver applies for a plate at random from the with-repetition pool in (a). What is the probability that the plate has no repeated letter and no repeated digit? Give an exact ratio and a decimal to three decimal places.驾驶员从 (a) 的"允许重复"池中随机申领一块车牌。求该车牌既无重复字母也无重复数字的概率。给出精确比值和保留三位小数的近似值。[3]
(d)If the province later allows plates of the form two letters, two digits, in any one of the two interleaved orders (i.e. LL-DD or LD-LD), how many plates are possible with the same $24$-letter, $9$-digit alphabets and repetition allowed? Justify your "multiply, not add" reasoning.若该省后续允许"两字母 + 两数字、两种交错顺序之一"的车牌（即 LL-DD 或 LD-LD），仍使用上述 $24$ 字母 / $9$ 数字字母表且允许重复，可生成多少种车牌？说明"相乘而非相加"的理由。[2]
Let $n$ be a positive integer. Consider the identity
$$ \sum_{r=0}^{n} \binom{n}{r} \;=\; 2^{n}. $$设 $n$ 为正整数。考察恒等式
$$ \sum_{r=0}^{n} \binom{n}{r} \;=\; 2^{n}. $$
(a)Derive this identity as a corollary of the Binomial Theorem by choosing a specific substitution into $(a + b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$. State the substitution and show the resulting line of algebra.在二项式定理 $(a + b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$ 中选取适当代入值，作为推论导出该恒等式。写出代入值并给出推导的代数行。[3]
(b)Give a combinatorial (double-counting) proof of the same identity. The left side counts what? The right side counts what? Explain in two sentences why both expressions count the same thing.给出同一恒等式的组合（双重计数）证明。左边在数什么？右边在数什么？用两句话解释为什么两边计数对象一致。[3]
(c)Verify the identity numerically for $n = 6$ by adding row $6$ of Pascal's triangle ($1, 6, 15, 20, 15, 6, 1$) and confirming that the sum equals $2^{6} = 64$.在 $n = 6$ 时数值验证：将帕斯卡三角形第 $6$ 行各项 ($1, 6, 15, 20, 15, 6, 1$) 相加，确认其和等于 $2^{6} = 64$。[2]
(d)Use the analogous substitution $a = 1$, $b = -1$ into the Binomial Theorem to derive the alternating-sum identity $\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} = 0$ (for $n \ge 1$). State the substitution; show the resulting line.类似地在二项式定理中令 $a = 1$、$b = -1$，导出交错和恒等式 $\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} = 0$（$n \ge 1$）。写出代入值并展示推导的代数行。[2]
Q12HARDHonors🇺🇸 US🇨🇦 ABAP-feeder FRQ§3 + §5 Capstone · HSS-CP.B.9 (+) / AB Math 30-1 SO 3, 4[9 marks]
A standard $52$-card deck is shuffled and a $5$-card poker hand is dealt at random (order does not matter).将一副标准 $52$ 张扑克牌洗匀，随机发出 $5$ 张牌组成一手（顺序不计）。
(a)State the size of the sample space, $\binom{52}{5}$, as a simplified integer.写出样本空间大小 $\binom{52}{5}$，化简为整数。[2]
(b)A "flush" is a $5$-card hand whose cards all share the same suit (there are $4$ suits, each of $13$ cards). Count the number of flushes."同花"是指 $5$ 张牌花色全相同的一手牌（共 $4$ 种花色，每种 $13$ 张）。求同花的数目。[3]
(c)Compute the probability of being dealt a flush, $P(\text{flush})$, as an exact ratio of binomial coefficients and as a decimal to four decimal places.求拿到同花的概率 $P(\text{同花})$，以二项式系数比的精确形式以及保留四位小数的近似值给出。[2]
(d)Independently, consider the expansion of $(x + 1)^{52}$. State the coefficient of $x^{5}$ in this expansion, and explain in one sentence why this number is equal to your answer in (a). Reference the combinatorial interpretation $\binom{n}{r}$ counts $r$-subsets of an $n$-set.另行考察 $(x + 1)^{52}$ 的展开。写出该展开式中 $x^{5}$ 的系数，并用一句话解释为何此数等于 (a) 中的答案。请引用组合学解释：$\binom{n}{r}$ 计数 $n$ 元集合的 $r$ 元子集。[2]
🇺🇸 US Common Core美国共同核心HSS-CP.B.9 (+) · HSA-APR.C.5 (+) · both plus-cluster (honors / Pre-Calc / AP-feeder); CCSSM does not host combinatorics in standard Algebra II两条均属 plus 簇（荣誉 / Pre-Calc / AP 衔接）；CCSSM 标准 Algebra II 不包含组合学
🇨🇦 British Columbia不列颠哥伦比亚PC 12 content topic "Permutations, Combinations, and Binomial Theorem" · counting principle, ${}_{n}P_{r}$, ${}_{n}C_{r}$, Pascal's triangle, $(a+b)^n$ for natural $n$PC 12 内容主题 "排列、组合与二项式定理" · 计数原理、${}_{n}P_{r}$、${}_{n}C_{r}$、帕斯卡三角形、自然数指数下的 $(a+b)^n$
🇨🇦 Alberta阿尔伯塔Math 30-1 General Outcome "Permutations, Combinations and Binomial Theorem" · SO 1 (counting) · SO 2 (${}_{n}P_{r}$) · SO 3 (${}_{n}C_{r}$) · SO 4 (Binomial Theorem, specific term)Math 30-1 总目标 "排列、组合与二项式定理" · SO 1（计数）· SO 2（${}_{n}P_{r}$）· SO 3（${}_{n}C_{r}$）· SO 4（二项式定理、指定项）
Full 4-column Syllabus Map lives in ../Study Guides/Unit_11_Combinatorics_and_the_Binomial_Theorem.html. US students: treat the whole unit as Honors / AP-feeder; every question carries an Honors chip when stamped US.完整 4 栏大纲对照见 ../Study Guides/Unit_11_Combinatorics_and_the_Binomial_Theorem.html。美国学生：整单元视为荣誉级 / AP 衔接；凡标注 US 的题目均加注 Honors 标签。