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Combinatorics and the Binomial Theorem · Solutions组合学与二项式定理 · 解析

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-feeder / ON / BC / AB styles练习题配套解析 · 逐分讲解 · SAT / AP 衔接 / ON / BC / AB 卷型

EASY MEDIUM HARD 🇺🇸 US 🇨🇦 ON 🇨🇦 BC 🇨🇦 AB SAT-style MCQ AP-feeder FRQ ON Provincial-style BC Provincial-style AB Provincial-style Honors
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  简答题 · 解析SAT MCQ + Provincial short answer · 18 marksSAT 选择题 + 省考简答题 · 18 分

Section A · Worked SolutionsA 节 · 详解

Q1EASYHonors 🇺🇸 US SAT-style MCQ §1 Counting Principle · HSS-CP.B.9 (+) [3 marks]

Cafeteria: $4$ sandwiches $\times$ $3$ sides $\times$ $5$ drinks. Distinct meals?自助餐厅：$4$ 种三明治 $\times$ $3$ 种配菜 $\times$ $5$ 种饮品。不同套餐数？

Answer:答案：  (B)  $60$ meals$60$ 种套餐

(a) Apply the fundamental counting principle应用基本计数原理 M1·A1·A1

A meal is built by three independent successive choices (sandwich, then side, then drink). The fundamental counting principle (multiplication rule) gives一份套餐由三次彼此独立的连续选择构成（先三明治，再配菜，再饮品）。由基本计数原理（乘法原理）：

$$ N \;=\; 4 \times 3 \times 5 \;=\; 60. $$

So there are $60$ distinct meals, matching option (B).故共 $60$ 种不同套餐，与选项 (B) 相符。

Why the wrong choices fail.错项分析。
  • (A) $12$, multiplies only $4 \times 3$ (forgets the drinks slot entirely).只乘 $4 \times 3$（完全漏掉了饮品一栏）。
  • (C) $120$, adds an extra factor of $2$ (perhaps mis-counting one slot, or treating a single choice as two).多乘了一个因子 $2$（可能某一栏数重了，或把一次选择算成了两次）。
  • (D) $720$, computes $4! \cdot 3! \cdot 5! / \text{something}$, or applies ${}_{n}P_{r}$ where no ordering is needed; classic over-count from reaching for permutations when only the multiplication rule is needed.计算了类似 $4! \cdot 3! \cdot 5! / \text{某式}$，或在无须考虑顺序时套用 ${}_{n}P_{r}$；典型的"只需乘法原理却动用排列"的多算错误。
If the choices are independent and "with replacement" (each slot draws from its own pool), just multiply pool sizes.若各选择相互独立且"放回式"（每栏从各自池中取），直接将各池大小相乘即可。 No factorials, no ${}_{n}P_{r}$, no $\binom{n}{r}$ are needed here, because the three pools (sandwich, side, drink) are different sets, not subsets of one set. Reach for factorials only when you are arranging or sub-selecting from the same pool. This is the single most common combinatorics misstep at the entry level: students who have just learnt permutations try to apply ${}_{n}P_{r}$ to multi-pool problems.此处既不需要阶乘，也不需要 ${}_{n}P_{r}$ 或 $\binom{n}{r}$，因为三个池（三明治、配菜、饮品）是不同的集合，并非同一集合的子集。只有在同一池中进行排列或子集选择时，才需要用到阶乘。这是入门阶段最常见的组合学误区：刚学完排列的学生总想把 ${}_{n}P_{r}$ 套到多池问题上。
Q2EASYHonors 🇺🇸 US SAT-style MCQ §2 Permutations · HSS-CP.B.9 (+) [3 marks]

Evaluate ${}_{7}P_{3}$.求 ${}_{7}P_{3}$ 的值。

Answer:答案：  (C)  $210$

(a) Apply the permutation formula套用排列公式 M1·A1

$$ {}_{7}P_{3} \;=\; \frac{7!}{(7-3)!} \;=\; \frac{7!}{4!} \;=\; 7 \cdot 6 \cdot 5 \;=\; 210. $$

(b) Cross-check by the "slots" intuition用"填位"直觉交叉验证 A1

${}_{7}P_{3}$ counts ordered arrangements of $3$ items drawn from $7$. Slot $1$: $7$ choices; slot $2$: $6$ remaining; slot $3$: $5$ remaining. Product $= 7 \cdot 6 \cdot 5 = 210$. $\checkmark$ Matches option (C).${}_{7}P_{3}$ 表示从 $7$ 项中有序选 $3$ 项的排列数。第 $1$ 位：$7$ 种选择；第 $2$ 位：剩 $6$；第 $3$ 位：剩 $5$。乘积 $= 7 \cdot 6 \cdot 5 = 210$。$\checkmark$ 与选项 (C) 相符。

Why the wrong choices fail.错项分析。
  • (A) $21$, $\binom{7}{2} = 21$, or $\binom{7}{5} = 21$; the student computed a combination instead of a permutation, possibly with the wrong $r$.$\binom{7}{2} = 21$ 或 $\binom{7}{5} = 21$；学生算的是组合而非排列，且可能 $r$ 也用错。
  • (B) $35$, this is $\binom{7}{3} = 35$, the combination value, which is ${}_{7}P_{3} / 3!$ — off by a factor of $3! = 6$ because order was discounted.即 $\binom{7}{3} = 35$，是组合值，等于 ${}_{7}P_{3} / 3!$，因忽略顺序而少了 $3! = 6$ 这一因子。
  • (D) $5040$, this is $7! = 5040$, the count of full arrangements of all $7$ items; the student forgot to stop after $3$ slots.即 $7! = 5040$，是 $7$ 个元素的全排列数；学生忘了取到第 $3$ 位就停。
${}_{n}P_{r} = n \cdot (n-1) \cdots (n-r+1)$, a product of exactly $r$ descending factors.${}_{n}P_{r} = n \cdot (n-1) \cdots (n-r+1)$，恰好是 $r$ 个递减因子的乘积。 The factorial-quotient definition $n!/(n-r)!$ is conceptually clean but slow; the slot-product version is faster. The key check: count the factors. For ${}_{7}P_{3}$ you write three factors $7 \cdot 6 \cdot 5$, not four ($7 \cdot 6 \cdot 5 \cdot 4 = 840$, a common slip) and not two ($7 \cdot 6 = 42$, the other common slip). The number of factors equals the number of slots, not the size of the pool.阶乘商定义 $n!/(n-r)!$ 概念清楚但慢；逐位相乘更快。关键检查：数因子个数。${}_{7}P_{3}$ 应写三个因子 $7 \cdot 6 \cdot 5$，不是四个（$7 \cdot 6 \cdot 5 \cdot 4 = 840$，常见误写）也不是两个（$7 \cdot 6 = 42$，另一种常见误写）。因子个数等于位数，而不是池的大小。
Q3MEDIUMHonors 🇺🇸 US SAT-style MCQ §4 Pascal Symmetry · HSA-APR.C.5 (+) [3 marks]

Which is equal to $\binom{15}{4}$?下列哪一项等于 $\binom{15}{4}$？

Answer:答案：  (B)  $\binom{15}{11}$

(a) Apply Pascal symmetry $\binom{n}{r} = \binom{n}{n-r}$应用帕斯卡对称性 $\binom{n}{r} = \binom{n}{n-r}$ M1·A1·A1

With $n = 15$, $r = 4$: the complementary index is $n - r = 15 - 4 = 11$. So取 $n = 15$、$r = 4$：互补指标为 $n - r = 15 - 4 = 11$。故

$$ \binom{15}{4} \;=\; \binom{15}{15-4} \;=\; \binom{15}{11}. $$

This matches option (B). (Numerical check: $\binom{15}{4} = \frac{15 \cdot 14 \cdot 13 \cdot 12}{24} = \frac{32760}{24} = 1365$, and $\binom{15}{11} = \binom{15}{4} = 1365$ by the same symmetry. $\checkmark$)与选项 (B) 相符。（数值核对：$\binom{15}{4} = \frac{15 \cdot 14 \cdot 13 \cdot 12}{24} = \frac{32760}{24} = 1365$，由同一对称性可知 $\binom{15}{11} = \binom{15}{4} = 1365$。$\checkmark$）

Why the wrong choices fail.错项分析。
  • (A) $\binom{15}{3}$, the student "decreased $r$ by $1$" instead of replacing $r$ by $n - r$; $\binom{15}{3} = 455 \ne 1365$.学生把 $r$ "减 $1$"，而非用 $n - r$ 代替；$\binom{15}{3} = 455 \ne 1365$。
  • (C) $\binom{4}{15}$, the indices are illegally swapped, $\binom{n}{r}$ requires $0 \le r \le n$, so $\binom{4}{15}$ is conventionally zero (no way to choose $15$ from $4$).非法地交换了上下指标；$\binom{n}{r}$ 要求 $0 \le r \le n$，按约定 $\binom{4}{15} = 0$（无法从 $4$ 个中取 $15$ 个）。
  • (D) $\binom{11}{4}$, the student replaced $n$ by $n - r$ but left $r$ alone; $\binom{11}{4} = 330 \ne 1365$.学生只把 $n$ 换成了 $n - r$，却没动 $r$；$\binom{11}{4} = 330 \ne 1365$。
Pascal symmetry is a swap of $r$ with $n - r$, the $n$ stays.帕斯卡对称性是将 $r$ 换成 $n - r$，$n$ 不变。 The combinatorial reason: choosing an $r$-subset of an $n$-set is the same as choosing the $(n - r)$-element complement to leave behind. Every $r$-subset $S$ pairs bijectively with its complement $S^{c}$, so the count is the same. AP-feeder problems love this swap because it converts an awkward $\binom{20}{17}$ into a fast $\binom{20}{3}$, three factors instead of seventeen. The diagnostic: whichever of $r$ and $n - r$ is smaller is the side you should compute on.组合学解释：从 $n$ 元集合中选 $r$ 元子集，等同于选 $n - r$ 元的留下子集（补集）。每个 $r$ 元子集 $S$ 与其补集 $S^{c}$ 一一对应，故计数相同。AP 衔接题尤爱此变换，能把笨重的 $\binom{20}{17}$ 转成 $\binom{20}{3}$——三个因子而非十七个。判定方法：$r$ 与 $n - r$ 中较小者，正是该实际计算的一侧。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §3 Combinations · MDM4U Counting and Probability [4 marks]

$12$ debate-club members; choose $3$. (a) Undifferentiated team count + justify $\binom{n}{r}$ vs ${}_{n}P_{r}$. (b) Distinct roles count. (c) Verify (b) $= 3! \times$ (a).辩论俱乐部 $12$ 人，选 $3$ 人。(a) 不区分角色的团队选法 + 说明用 $\binom{n}{r}$ 还是 ${}_{n}P_{r}$。(b) 分配三个不同角色的方法数。(c) 验证 (b) $= 3! \times$ (a)。

Answer:答案：  (a) $\binom{12}{3} = 220$ teams (order does not matter)$\binom{12}{3} = 220$ 个团队（顺序不计）  ·  (b) ${}_{12}P_{3} = 1320$ role-assignments${}_{12}P_{3} = 1320$ 种角色安排  ·  (c) $1320 = 3! \cdot 220$, factor of $3! = 6$ counts the role-permutations of any fixed $3$-team$1320 = 3! \cdot 220$，$3! = 6$ 计的是任一固定 $3$ 人团队的角色排列数

(a) Undifferentiated team: order does not matter不区分角色的团队：顺序不计 M1·A1

Sending three people with no roles means the order of selection is irrelevant. Use combinations, $\binom{n}{r}$:派三人但不分角色，意味着选人的顺序无关紧要。用组合 $\binom{n}{r}$：

$$ \binom{12}{3} \;=\; \frac{12!}{3! \cdot 9!} \;=\; \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \;=\; \frac{1320}{6} \;=\; 220. $$

Justification: "the three attendees are sent as an undifferentiated team", so swapping who-was-listed-first does not produce a new outcome.理由："三位代表作为不区分角色的团队前往"，故交换名单顺序不会产生新方案。

(b) Distinct roles: order matters不同角色：顺序有关 M1

Assigning the three roles captain, first speaker, second speaker distinguishes the three attendees, so order matters. Use permutations:分配队长、一辩、二辩三个不同角色后，三人有区分，故顺序有关。用排列：

$$ {}_{12}P_{3} \;=\; 12 \cdot 11 \cdot 10 \;=\; 1320. $$

(c) Verify and explain the factor of $3!$验证并解释 $3!$ 这一因子 R1

Numerical: $3! \cdot 220 = 6 \cdot 220 = 1320$. $\checkmark$
Combinatorial: every unordered $3$-team in (a) can be assigned the three distinct roles in $3! = 6$ ways (pick a captain in $3$ ways, then a first speaker in $2$, then second is forced). So ${}_{12}P_{3} = \binom{12}{3} \cdot 3!$, i.e. permutations count combinations multiplied by the role-permutations of each fixed team.数值验证：$3! \cdot 220 = 6 \cdot 220 = 1320$。$\checkmark$
组合学解释：(a) 中每个无序 $3$ 人团队都可用 $3! = 6$ 种方式分配三个不同角色（先选队长有 $3$ 种，再选一辩有 $2$ 种，二辩自动确定）。故 ${}_{12}P_{3} = \binom{12}{3} \cdot 3!$，即排列数等于组合数乘以每个固定团队的角色排列数。

Permutations $= $ combinations $\times $ (orderings of each subset): ${}_{n}P_{r} = \binom{n}{r} \cdot r!$.排列 $= $ 组合 $\times $（每个子集的排序数）：${}_{n}P_{r} = \binom{n}{r} \cdot r!$。 This single identity is the structural backbone of the unit. Read in either direction: divide ${}_{n}P_{r}$ by $r!$ to "forget order" and land in combinations; multiply $\binom{n}{r}$ by $r!$ to "decorate" each subset with a role-assignment and land in permutations. The MDM4U marker rewards students who explicitly cite the order-matters/order-does-not test before writing any number, the M1 is given for naming the right tool, not for the arithmetic.这条单一恒等式是整单元的结构主干。双向理解：${}_{n}P_{r} \div r!$ 即"忘掉顺序"得到组合；$\binom{n}{r} \times r!$ 即"为每个子集贴上角色"得到排列。MDM4U 阅卷者偏爱在写任何数字之前就明确"是否计较顺序"的学生——M1 给的是"叫对工具"，而不是算对数。
Q5MEDIUM 🇨🇦 AB AB Provincial-style §2 Identical-Elements Permutations · AB Math 30-1 SO 2 (indicator 2.6) [5 marks]

Word STATISTICS ($10$ letters: $3$ S, $3$ T, $1$ A, $2$ I, $1$ C). (a) State formula. (b) Count arrangements. (c) Repeat for STATS.单词 STATISTICS（$10$ 字母：$3$ S、$3$ T、$1$ A、$2$ I、$1$ C）。(a) 写出公式。(b) 数排列总数。(c) 对 STATS 重复。

Answer:答案：  (a) $\dfrac{n!}{k_{1}!\, k_{2}!\, \cdots k_{m}!}$  ·  (b) $50{,}400$ arrangements$50{,}400$ 种排列  ·  (c) $30$ arrangements$30$ 种排列

(a) Identical-elements formula相同元素公式 A1

For a word of length $n$ with letter multiplicities $k_{1}, k_{2}, \ldots, k_{m}$ (summing to $n$), the number of distinguishable arrangements is设单词长度为 $n$，各字母重数为 $k_{1}, k_{2}, \ldots, k_{m}$（和为 $n$），则可区分排列数为

$$ N \;=\; \frac{n!}{k_{1}!\, k_{2}!\, \cdots k_{m}!}. $$

The $n!$ in the numerator counts all arrangements as if the letters were distinct; each $k_{i}!$ in the denominator divides out the permutations among identical copies of letter $i$, which produce the same word.分子 $n!$ 假设所有字母互异时的全部排列；分母每个 $k_{i}!$ 除掉第 $i$ 种字母的相同副本间所产生的等价排列。

(b) Apply to STATISTICS代入 STATISTICS M1·A1

Multiplicities: S$=3$, T$=3$, A$=1$, I$=2$, C$=1$. Total $n = 3 + 3 + 1 + 2 + 1 = 10$. $\checkmark$各重数：S$=3$、T$=3$、A$=1$、I$=2$、C$=1$。总数 $n = 3 + 3 + 1 + 2 + 1 = 10$。$\checkmark$

$$ N \;=\; \frac{10!}{3!\, 3!\, 1!\, 2!\, 1!} \;=\; \frac{3{,}628{,}800}{6 \cdot 6 \cdot 1 \cdot 2 \cdot 1} \;=\; \frac{3{,}628{,}800}{72} \;=\; 50{,}400. $$

(c) Apply to STATS代入 STATS A1·A1

Multiplicities: S$=2$, T$=2$, A$=1$. Total $n = 5$. $\checkmark$各重数：S$=2$、T$=2$、A$=1$。总数 $n = 5$。$\checkmark$

$$ N \;=\; \frac{5!}{2!\, 2!\, 1!} \;=\; \frac{120}{4} \;=\; 30. $$
The denominator is one factorial per letter type, not per repeated letter.分母按字母种类每种一个阶乘，而非按"每个重复字母"。 Common slip: students see "three S's and three T's" and write $\tfrac{10!}{(3 \cdot 3)!} = \tfrac{10!}{9!} = 10$, treating "the six repeated letters" as a single block. The denominator is a product of factorials, one per distinct letter, $3! \cdot 3! \cdot 1! \cdot 2! \cdot 1!$, never $(3 \cdot 3 \cdot 1 \cdot 2 \cdot 1)!$ nor $(3 + 3 + 1 + 2 + 1)!$. AB Math 30-1 markers explicitly check that you have listed one factorial per letter type; missing the $1!$s costs no marks (since $1! = 1$) but writing them shows the audit was done.常见误写：学生见"三个 S 与三个 T"便写成 $\tfrac{10!}{(3 \cdot 3)!} = \tfrac{10!}{9!} = 10$，把"六个重复字母"当作一个整体。分母应是乘积形式，每种不同字母一个阶乘：$3! \cdot 3! \cdot 1! \cdot 2! \cdot 1!$，绝不是 $(3 \cdot 3 \cdot 1 \cdot 2 \cdot 1)!$ 也不是 $(3 + 3 + 1 + 2 + 1)!$。AB Math 30-1 阅卷者明确检查"每种字母列一个阶乘"；漏写 $1!$ 不扣分（因 $1! = 1$），但写出来证明你审过题。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  长答题 · 解析AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 35 分

Section B · Worked SolutionsB 节 · 详解

Q6MEDIUM 🇨🇦 BC BC Provincial-style §3 Committee Counts · BC PC 12 Permutations/Combinations [8 marks]

Choose $5$ from $8$ women + $6$ men. (a) No restriction. (b) Exactly $3$W $+ 2$M. (c) At least $4$W.从 $8$ 女 + $6$ 男中选 $5$ 人。(a) 无限制。(b) 恰好 $3$ 女 + $2$ 男。(c) 至少 $4$ 女。

Answer:答案：  (a) $\binom{14}{5} = 2{,}002$  ·  (b) $\binom{8}{3}\binom{6}{2} = 840$  ·  (c) $\binom{8}{4}\binom{6}{1} + \binom{8}{5}\binom{6}{0} = 476$

(a) Unrestricted committee无限制委员会 M1·A1

The pool is $8 + 6 = 14$ people and we choose $5$ unordered (a committee has no roles in this part):总池为 $8 + 6 = 14$ 人，无序选 $5$ 人（本小题委员会不分角色）：

$$ \binom{14}{5} \;=\; \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{5!} \;=\; \frac{240{,}240}{120} \;=\; 2{,}002. $$

(b) Exactly $3$ women and $2$ men恰好 $3$ 女 $2$ 男 M1·A1·A1

Split the selection: choose $3$ women from $8$, and independently choose $2$ men from $6$. By the multiplication rule:把选拔分两步：从 $8$ 女中选 $3$ 人，同时独立地从 $6$ 男中选 $2$ 人。由乘法原理：

$$ \binom{8}{3} \cdot \binom{6}{2} \;=\; 56 \cdot 15 \;=\; 840. $$

(c) At least $4$ women至少 $4$ 女 M1·A1·A1

"At least $4$ women" out of $5$ committee members partitions into two disjoint cases: exactly $4$W$+1$M, or exactly $5$W$+0$M. Add the case counts:$5$ 人委员会中"至少 $4$ 女"可分为两个互不相交的情形：恰好 $4$ 女 $+ 1$ 男，或恰好 $5$ 女 $+ 0$ 男。两种情况相加：

$$ \binom{8}{4} \cdot \binom{6}{1} \;+\; \binom{8}{5} \cdot \binom{6}{0} \;=\; 70 \cdot 6 \;+\; 56 \cdot 1 \;=\; 420 + 56 \;=\; 476. $$
"And" $\Rightarrow $ multiply within a case; "or" $\Rightarrow $ add across disjoint cases."且" $\Rightarrow $ 同一情形内相乘；"或" $\Rightarrow $ 不相交情形间相加。 The two-rule split is the spine of every committee/lineup problem. "Exactly $3$W and $2$M" is one case, so multiply $\binom{8}{3} \cdot \binom{6}{2}$. "At least $4$W" is the union of two disjoint cases ($4$W+$1$M, $5$W+$0$M), so add their counts; the two cases are disjoint (no committee has both exactly $4$ and exactly $5$ women), so no inclusion-exclusion adjustment is needed. BC PC 12 markers explicitly look for the case split written out before any binomial coefficients are computed.这一"两规则"区分是所有委员会 / 排队题的骨架。"恰好 $3$ 女且 $2$ 男"是单一情形，故相乘 $\binom{8}{3} \cdot \binom{6}{2}$。"至少 $4$ 女"是两不相交情形（$4$ 女 $1$ 男、$5$ 女 $0$ 男）之并，故相加；两情形不相交（同一委员会不可能既恰好 $4$ 女又恰好 $5$ 女），无需作容斥调整。BC PC 12 阅卷者明确要求在写任何二项式系数之前先把情形分拆写出。
Q7MEDIUM 🇨🇦 AB AB Provincial-style §5 Binomial Theorem · AB Math 30-1 SO 4 (indicators 4.1, 4.5) [8 marks]

$(2x + 3)^{5}$. (a) Row $5$. (b) General term $T_{r+1}$. (c) Full expansion. (d) Sum of coefficients.$(2x + 3)^{5}$。(a) 第 $5$ 行。(b) 通项 $T_{r+1}$。(c) 完整展开。(d) 系数之和。

Answer:答案：  (a) $1, 5, 10, 10, 5, 1$  ·  (c) $32x^{5} + 240x^{4} + 720x^{3} + 1080x^{2} + 810x + 243$  ·  (d) $3125 = 5^{5}$ $\checkmark$

(a) Row $5$ of Pascal's triangle帕斯卡三角形第 $5$ 行 A1

Row $5$ is $\binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5}$, which gives第 $5$ 行为 $\binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5}$，即

$$ 1, \;5, \;10, \;10, \;5, \;1. $$

Symmetric, as Pascal's rule and the $\binom{n}{r} = \binom{n}{n-r}$ identity predict.关于中线对称，恰如帕斯卡法则与 $\binom{n}{r} = \binom{n}{n-r}$ 所预言。

(b) General term通项 M1·A1

With $a = 2x$, $b = 3$, $n = 5$:取 $a = 2x$、$b = 3$、$n = 5$：

$$ T_{r+1} \;=\; \binom{5}{r} (2x)^{5-r} (3)^{r} \;=\; \binom{5}{r} \cdot 2^{5-r} \cdot 3^{r} \cdot x^{5-r}, \qquad r \in \{0, 1, 2, 3, 4, 5\}. $$

(c) Full expansion完整展开 M1·A1·A1·A1

Tabulate $T_{r+1}$ for each $r$:逐个 $r$ 列出 $T_{r+1}$：

  • $r = 0$: $\binom{5}{0} \cdot 2^{5} \cdot 3^{0} \cdot x^{5} = 1 \cdot 32 \cdot 1 \cdot x^{5} = 32 x^{5}$.
  • $r = 1$: $\binom{5}{1} \cdot 2^{4} \cdot 3^{1} \cdot x^{4} = 5 \cdot 16 \cdot 3 \cdot x^{4} = 240 x^{4}$.
  • $r = 2$: $\binom{5}{2} \cdot 2^{3} \cdot 3^{2} \cdot x^{3} = 10 \cdot 8 \cdot 9 \cdot x^{3} = 720 x^{3}$.
  • $r = 3$: $\binom{5}{3} \cdot 2^{2} \cdot 3^{3} \cdot x^{2} = 10 \cdot 4 \cdot 27 \cdot x^{2} = 1080 x^{2}$.
  • $r = 4$: $\binom{5}{4} \cdot 2^{1} \cdot 3^{4} \cdot x^{1} = 5 \cdot 2 \cdot 81 \cdot x = 810 x$.
  • $r = 5$: $\binom{5}{5} \cdot 2^{0} \cdot 3^{5} \cdot x^{0} = 1 \cdot 1 \cdot 243 = 243$.

Sum:求和：

$$ (2x + 3)^{5} \;=\; 32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243. $$

(d) Sum-of-coefficients check系数之和的验算 A1

Sum of coefficients $= 32 + 240 + 720 + 1080 + 810 + 243 = 3125$. Plugging $x = 1$ into the original: $(2(1) + 3)^{5} = 5^{5} = 3125$. $\checkmark$ Agreement confirms the expansion.系数之和 $= 32 + 240 + 720 + 1080 + 810 + 243 = 3125$。在原式中代入 $x = 1$：$(2(1) + 3)^{5} = 5^{5} = 3125$。$\checkmark$ 两边一致，展开正确。

The $x = 1$ trick instantly audits any binomial expansion.$x = 1$ 法可瞬时验算任何二项展开。 For any polynomial $p(x)$, $p(1)$ equals the sum of its coefficients. So plugging $x = 1$ into both the original $(2x+3)^{5}$ and your expanded sum gives a one-line consistency check: $5^{5} = 3125$ must equal $\sum c_{i}$. Likewise $p(-1) = $ alternating sum of coefficients, useful for verifying signs in $(a - b)^{n}$ problems. AB Math 30-1 markers explicitly reward students who include the $x = 1$ check, it costs one line and catches every arithmetic slip.对任意多项式 $p(x)$，$p(1)$ 即所有系数之和。故在原式 $(2x+3)^{5}$ 与你写出的展开式中同时代入 $x = 1$，便有一行一致性验算：$5^{5} = 3125$ 必须等于 $\sum c_{i}$。类似地，$p(-1) = $ 系数交错和，可用于 $(a - b)^{n}$ 题中验证符号。AB Math 30-1 阅卷者明确鼓励学生加上 $x = 1$ 验算——只需一行，却能截获所有算术失误。
Q8HARD 🇨🇦 ON ON Provincial-style §6 Term Extraction (positive powers) · MHF4U Polynomial Functions [9 marks]

$(2x - 3)^{12}$. (a) $T_{r+1}$. (b) $r$ for $x^{7}$. (c) Coefficient of $x^{7}$. (d) Sign + reasoning.

Answer:  (a) $T_{r+1} = \binom{12}{r} 2^{12-r} (-3)^{r} x^{12-r}$  ·  (b) $r = 5$  ·  (c) $-24{,}634{,}368$  ·  (d) negative because $(-3)^{r}$ alternates and $r = 5$ is odd

(a) General term M1·A1·A1

With $a = 2x$, $b = -3$, $n = 12$: $$ T_{r+1} \;=\; \binom{12}{r} (2x)^{12-r} (-3)^{r} \;=\; \binom{12}{r} \cdot 2^{12-r} \cdot (-3)^{r} \cdot x^{12-r}, \qquad r \in \{0, 1, \ldots, 12\}. $$ The dependence on $x$ is the single power $x^{12-r}$.

(b) Solve for the $x^{7}$ index M1·A1

Match the exponent: $12 - r = 7 \Longrightarrow r = 5$.

(c) Coefficient at $r = 5$ M1·A1·A1

Substitute $r = 5$ into the coefficient (the part of $T_{r+1}$ not involving $x$): $$ c_{7} \;=\; \binom{12}{5} \cdot 2^{12-5} \cdot (-3)^{5} \;=\; 792 \cdot 2^{7} \cdot (-243) \;=\; 792 \cdot 128 \cdot (-243). $$ Compute step by step: $792 \cdot 128 = 101{,}376$; then $101{,}376 \cdot (-243) = -24{,}634{,}368$. So $$ c_{7} \;=\; -24{,}634{,}368. $$

(d) Sign and alternation A1

The coefficient is negative. Reason: every increment of $r$ by $1$ multiplies the coefficient by $-3/2 \cdot (12-r)/(r+1)$, picking up one extra factor of $-3$ each step. Since $r = 5$ is odd, $(-3)^{5} < 0$, so the entire coefficient inherits a negative sign. (Even $r$ produces a positive coefficient; odd $r$, negative.)
"Find the term containing $x^{k}$" $\Rightarrow $ solve $n - r = k$, then substitute. The general term in $(a x + b)^{n}$ has $x$-power $n - r$ (when $a$ is the linear part). So extracting the $x^{k}$ coefficient is two lines: solve the linear equation for $r$, then compute $\binom{n}{r} a^{n-r} b^{r}$. The MHF4U marker awards two distinct A1 marks: one for finding the right $r$ (the algebra step), one for the correct numerical coefficient including sign. Drop the sign and lose an A1 even if the magnitude is correct.
Q9HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §6 Term Extraction (negative powers) · BC PC 12 / HSA-APR.C.5 (+) [10 marks]

$\left(x + \tfrac{2}{x}\right)^{10}$, $x \ne 0$. (a) $T_{r+1}$. (b) $r$ for constant. (c) Constant term. (d) $r$ for $x^{4}$ + coefficient. (e) Why only even powers.

Answer:  (a) $T_{r+1} = \binom{10}{r} 2^{r} x^{10-2r}$  ·  (b) $r = 5$  ·  (c) $8{,}064$  ·  (d) $r = 3$, coefficient $960$  ·  (e) exponent $10 - 2r$ is always even

(a) General term M1·A1

With $a = x$, $b = \tfrac{2}{x} = 2 x^{-1}$, $n = 10$: $$ T_{r+1} \;=\; \binom{10}{r} x^{10-r} \cdot \left(\tfrac{2}{x}\right)^{r} \;=\; \binom{10}{r} x^{10-r} \cdot 2^{r} x^{-r} \;=\; \binom{10}{r} \cdot 2^{r} \cdot x^{10-2r}. $$ The $x$-exponent simplifies to $10 - 2r$ because the two contributions $x^{10-r}$ and $x^{-r}$ combine additively in the exponent.

(b) Constant-term index M1·A1

Constant $\Leftrightarrow x^{0}$ $\Leftrightarrow 10 - 2r = 0 \Leftrightarrow r = 5$.

(c) Constant term M1·A1·A1

Substitute $r = 5$ (and the $x$-part vanishes, since $x^{0} = 1$): $$ T_{6} \;=\; \binom{10}{5} \cdot 2^{5} \;=\; 252 \cdot 32 \;=\; 8{,}064. $$

(d) $x^{4}$ index and coefficient M1·A1

Solve $10 - 2r = 4 \Rightarrow r = 3$. Coefficient: $$ c_{4} \;=\; \binom{10}{3} \cdot 2^{3} \;=\; 120 \cdot 8 \;=\; 960. $$

(e) Why only even powers appear R1

The $x$-exponent in $T_{r+1}$ is $10 - 2r$, which is even for every integer $r \in \{0, 1, \ldots, 10\}$ (subtracting an even number from $10$ stays even). So the expansion has only even-power terms: $x^{10}, x^{8}, x^{6}, x^{4}, x^{2}, x^{0}, x^{-2}, x^{-4}, x^{-6}, x^{-8}, x^{-10}$. No odd-power terms can appear.
Negative-power expansions live entirely in the exponent algebra. The whole trick of $(x + b x^{-1})^{n}$ is that combining $x^{n-r}$ with $(x^{-1})^{r} = x^{-r}$ gives $x^{(n-r) + (-r)} = x^{n - 2r}$, exponent changes by $2$ for each unit step in $r$. So the family of exponents is an arithmetic progression of step $-2$. The "constant term exists $\Leftrightarrow n$ is even" diagnostic falls out immediately: $n - 2r = 0$ requires $r = n/2$, which is an integer only when $n$ is even. AP-feeder problems love this because it stress-tests both the Binomial Theorem and the algebra of exponents in one prompt.
PART III  ·  MODELING / IDENTITIES · SOLUTIONS第三部分  ·  建模 / 恒等式 · 解析Universal + honors capstone · 28 marks通用题 + 荣誉级压轴 · 28 分

Section C · Worked SolutionsC 节 · 详解

Q10MEDIUM 🇨🇦 ON ON Provincial-style §1 + §3 Modeling · MDM4U Counting and Probability [9 marks]

Plates LLL-DDD; $24$ usable letters, $9$ usable digits. (a) With repetition. (b) No repeated letter or digit. (c) Probability of (b) within (a). (d) Two interleaved orders LLDD, LDLD.

Answer:  (a) $10{,}077{,}696$ plates  ·  (b) $6{,}120{,}576$ plates  ·  (c) $\tfrac{6{,}120{,}576}{10{,}077{,}696} \approx 0.607$  ·  (d) $2 \cdot 24^{2} \cdot 9^{2} = 93{,}312$

(a) Plates with repetition M1·A1

Six independent slots, three from a $24$-letter pool and three from a $9$-digit pool, all with replacement: $$ N_{a} \;=\; 24^{3} \cdot 9^{3} \;=\; 13{,}824 \cdot 729 \;=\; 10{,}077{,}696. $$

(b) Plates with no repeat within letters or digits M1·A1

Letters: $24 \cdot 23 \cdot 22 = 12{,}144$ ordered triples (a permutation). Digits: $9 \cdot 8 \cdot 7 = 504$. Multiply: $$ N_{b} \;=\; 12{,}144 \cdot 504 \;=\; 6{,}120{,}576. $$

(c) Probability M1·A1·A1

Under the uniform-random-plate model of (a): $$ P \;=\; \frac{N_{b}}{N_{a}} \;=\; \frac{6{,}120{,}576}{10{,}077{,}696} \;=\; \frac{23 \cdot 22}{24 \cdot 24} \cdot \frac{8 \cdot 7}{9 \cdot 9} \;=\; \frac{506}{576} \cdot \frac{56}{81} \;=\; \frac{28{,}336}{46{,}656}. $$ Decimal: $P \approx 0.6073\ldots \approx 0.607$.

(d) Two interleaved orders, repetition allowed A1·A1

For each fixed ordering pattern (e.g. LLDD), the count is $24^{2} \cdot 9^{2} = 576 \cdot 81 = 46{,}656$. There are $2$ ordering patterns (LLDD and LDLD), and they are disjoint (no plate fits both patterns), so add: $$ N_{d} \;=\; 2 \cdot 46{,}656 \;=\; 93{,}312. $$ "Multiply, not add" inside each pattern (the four slots are independent choices, so we multiply); "add across patterns" because the patterns themselves are mutually exclusive (an LLDD-plate cannot also be an LDLD-plate). So the full formula is $\text{(slots multiplied)} \times \text{(patterns added)} = 2 \cdot 24^{2} \cdot 9^{2}$.
"With/without replacement" toggles between $n^{r}$ and $_{n}P_{r}$. The two facts to internalise: (i) ordered draws with replacement from a pool of size $n$, $r$ times, gives $n^{r}$, because each slot independently has $n$ choices; (ii) ordered draws without replacement give $_{n}P_{r} = n(n-1)\cdots(n-r+1)$. The license-plate framing is the textbook prompt to test this distinction: part (a) is $n^{r}$ per pool, part (b) is $_{n}P_{r}$ per pool. The ratio in (c) then becomes $\tfrac{_{n}P_{r}}{n^{r}}$, the "no-collision" probability that famously powers the birthday paradox at $r = 23$, $n = 365$. MDM4U markers explicitly reward students who recognise this structural connection.
Q11HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §7 Subset-Count Identities · HSA-APR.C.5 (+) [10 marks]

$\sum_{r=0}^{n} \binom{n}{r} = 2^{n}$. (a) Derive via $(a+b)^{n}$. (b) Combinatorial proof. (c) Numerical $n=6$. (d) Alternating-sum identity via $a=1, b=-1$.

Answer:  (a) Substitute $a = b = 1$  ·  (b) both sides count subsets of $\{1, \ldots, n\}$  ·  (c) $1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^{6}$ $\checkmark$  ·  (d) $0^{n} = 0$ for $n \ge 1$

(a) Binomial-Theorem derivation M1·A1·A1

Substitute $a = 1$, $b = 1$ into the Binomial Theorem: $$ (1 + 1)^{n} \;=\; \sum_{r=0}^{n} \binom{n}{r} \cdot 1^{n-r} \cdot 1^{r} \;=\; \sum_{r=0}^{n} \binom{n}{r}. $$ But $(1 + 1)^{n} = 2^{n}$ directly, so $\sum_{r=0}^{n} \binom{n}{r} = 2^{n}$. $\;\blacksquare$ AG

(b) Combinatorial (double-counting) proof M1·A1·A1

Let $S = \{1, 2, \ldots, n\}$. Count the number of subsets of $S$ in two ways:
  • Left side. Partition the subsets by size: there are $\binom{n}{r}$ subsets of size $r$. Summing across all sizes $r = 0, 1, \ldots, n$ gives the total count $\sum_{r=0}^{n} \binom{n}{r}$.
  • Right side. Build a subset by making $n$ independent yes/no decisions (one per element of $S$): "is element $i$ in the subset?" Each decision has $2$ outcomes, so there are $2^{n}$ subsets total.
Since both sides count the same set of objects (all subsets of $S$), the identity follows.

(c) Numerical check at $n = 6$ A1·A1

Row $6$: $1, 6, 15, 20, 15, 6, 1$. Sum: $$ 1 + 6 + 15 + 20 + 15 + 6 + 1 \;=\; 64 \;=\; 2^{6}. \; \checkmark $$

(d) Alternating-sum identity via $a = 1, b = -1$ M1·A1

Substitute $a = 1$, $b = -1$ into the Binomial Theorem: $$ (1 + (-1))^{n} \;=\; \sum_{r=0}^{n} \binom{n}{r} \cdot 1^{n-r} \cdot (-1)^{r} \;=\; \sum_{r=0}^{n} (-1)^{r} \binom{n}{r}. $$ The left side is $0^{n} = 0$ for every $n \ge 1$ (the case $n = 0$ gives $0^{0}$, which is conventionally $1$, and is excluded by the $n \ge 1$ hypothesis). So $\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} = 0$ for $n \ge 1$. $\;\blacksquare$
The Binomial Theorem is an identity machine, choose values of $a, b$ and read off corollaries. Substituting $(a, b) = (1, 1)$ extracts the sum of every row of Pascal's triangle ($2^{n}$); $(1, -1)$ extracts the alternating sum ($0$ for $n \ge 1$); $(1, x)$ for arbitrary $x$ gives a polynomial identity, $(1, 2)$ gives a $3^{n}$ identity; etc. The double-counting proof in (b) is the deeper view: every binomial identity has a combinatorial interpretation, the algebraic derivation is one route, the bijective/double-count argument is the other. AP-feeder problems explicitly ask for "two proofs", testing both literacies. The $2^{n}$ identity is also the simplest example of a generating-function evaluation, which is the bridge into discrete math and combinatorics at the university level.
Q12HARDHonors 🇺🇸 US 🇨🇦 AB AP-feeder FRQ §3 + §5 Capstone · HSS-CP.B.9 (+) / AB Math 30-1 SO 3, 4 [9 marks]

$5$-card poker hand from $52$-card deck. (a) $\binom{52}{5}$ as integer. (b) Number of flushes. (c) $P(\text{flush})$ exact + decimal. (d) Coefficient of $x^{5}$ in $(x+1)^{52}$ + interpretation.

Answer:  (a) $\binom{52}{5} = 2{,}598{,}960$  ·  (b) $4 \cdot \binom{13}{5} = 5{,}148$  ·  (c) $\tfrac{5148}{2598960} \approx 0.0020$  ·  (d) coefficient $= \binom{52}{5} = 2{,}598{,}960$

(a) Sample-space size M1·A1

A $5$-card hand drawn without order from a $52$-card deck: $$ \binom{52}{5} \;=\; \frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5!} \;=\; \frac{311{,}875{,}200}{120} \;=\; 2{,}598{,}960. $$

(b) Count of flushes M1·A1·A1

A flush is built in two stages: (1) choose the common suit ($4$ choices), and (2) choose $5$ cards from the $13$ cards of that suit ($\binom{13}{5} = 1{,}287$ choices). By the multiplication rule: $$ N_{\text{flush}} \;=\; 4 \cdot \binom{13}{5} \;=\; 4 \cdot 1{,}287 \;=\; 5{,}148. $$ (Note: this count includes straight flushes and royal flushes, which are usually treated as separate categories in poker rankings; the problem just asks for "flush" in the broad sense.)

(c) Probability A1·A1

$$ P(\text{flush}) \;=\; \frac{N_{\text{flush}}}{\binom{52}{5}} \;=\; \frac{4 \cdot \binom{13}{5}}{\binom{52}{5}} \;=\; \frac{5{,}148}{2{,}598{,}960}. $$ Decimal: $P \approx 0.001981\ldots \approx 0.0020$ to four decimal places (about $1$ flush every $505$ hands).

(d) Coefficient in $(x + 1)^{52}$ + combinatorial meaning M1·A1

By the Binomial Theorem, $$ (x + 1)^{52} \;=\; \sum_{r = 0}^{52} \binom{52}{r} x^{52 - r} \;=\; \sum_{k = 0}^{52} \binom{52}{52 - k} x^{k} \;=\; \sum_{k = 0}^{52} \binom{52}{k} x^{k}, $$ using Pascal symmetry $\binom{52}{52 - k} = \binom{52}{k}$. So the coefficient of $x^{5}$ is $\binom{52}{5} = 2{,}598{,}960$. This equals the sample-space size in (a): both count $5$-element subsets of a $52$-element set. Algebraically, $\binom{52}{5}$ counts the number of ways to choose $5$ "$x$"-factors out of $52$ when expanding $(x + 1)^{52}$ as a product; combinatorially, those $5$ chosen factors index a $5$-subset of the $52$ binomial factors, which is exactly the same counting structure as drawing $5$ cards from a $52$-card deck.
$\binom{n}{r}$ has three faces: algebraic, combinatorial, and generating-functional. The same number $\binom{52}{5} = 2{,}598{,}960$ shows up as: (i) the number of $5$-card poker hands (combinatorial), (ii) the coefficient of $x^{5}$ in $(x + 1)^{52}$ (algebraic / generating function), and (iii) the value of $\tfrac{52!}{5!\,47!}$ (factorial formula). The unification is the substance of the Binomial Theorem: $\binom{n}{r}$ is defined combinatorially, computed via the factorial formula, and read off the generating function. AP-feeder problems use this trifecta to bridge into Unit 13 (probability) and into the introductory combinatorics of first-year university discrete math. The deeper reason flushes are rare (~$0.2\%$): only one in $\binom{52}{5}/(4\binom{13}{5}) \approx 505$ hands has all five cards from the same suit, because most $5$-card subsets span multiple suits.