IB Math AA HL — Unit A3: Combinatorics — Solutions

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Section A — Worked SolutionsA 节 —— 详细解析

Q1MEDIUMPaper 1A1.9 Binomial — Term Hunt[6 marks]

Expansion of $\left(2x^{2} - \dfrac{1}{x}\right)^{9}$ — (a) general term, (b) coefficient of $x^{6}$, (c) term independent of $x$.$\left(2x^{2} - \dfrac{1}{x}\right)^{9}$ 的展开式 —— (a) 通项，(b) $x^{6}$ 的系数，(c) 不含 $x$ 的项。

Answers:答案： (a) $T_{r+1} = \binom{9}{r}\,2^{9-r}\,(-1)^{r}\,x^{18-3r}$ · (b) $4032$ · (c) $672$

(a) General term M1·A1

The binomial theorem gives $\displaystyle (a+b)^{n} = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}$. Apply with $a = 2x^{2}$, $b = -x^{-1}$, $n = 9$: $$ T_{r+1} \;=\; \binom{9}{r}\,(2x^{2})^{9-r}\,(-x^{-1})^{r} \;=\; \binom{9}{r}\,2^{\,9-r}\,(-1)^{r}\,x^{\,2(9-r)}\,x^{-r}. $$ Combine the $x$-powers: $x^{18-2r-r} = x^{18-3r}$. So $$ T_{r+1} = \binom{9}{r}\,2^{9-r}\,(-1)^{r}\,x^{18-3r}, \qquad r = 0, 1, \ldots, 9. $$

(b) Coefficient of $x^{6}$ M1·A1

Set the exponent to $6$: $18 - 3r = 6 \Rightarrow r = 4$. The $r = 4$ term is $$ \binom{9}{4}\,2^{\,5}\,(-1)^{4}\,x^{6} \;=\; 126 \cdot 32 \cdot 1 \cdot x^{6} \;=\; 4032\,x^{6}. $$ Coefficient $= \boxed{4032}$.

(c) Term independent of $x$ M1·A1

Set $18 - 3r = 0 \Rightarrow r = 6$. The $r = 6$ term is $$ \binom{9}{6}\,2^{\,3}\,(-1)^{6} \;=\; 84 \cdot 8 \cdot 1 \;=\; 672. $$ So the constant term in the expansion is $\boxed{672}$.

Trap to avoid: students often forget the $(-1)^{r}$ factor from the negative second term, or square the exponent of $x^{2}$ wrong (it's $2(9-r)$, not $(9-r)^{2}$). Always combine the two $x$-powers into a single exponent before hunting for the target $r$.

(a) 通项 M1·A1

二项定理（binomial theorem）给出 $\displaystyle (a+b)^{n} = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}$。取 $a = 2x^{2}$、$b = -x^{-1}$、$n = 9$： $$ T_{r+1} \;=\; \binom{9}{r}\,(2x^{2})^{9-r}\,(-x^{-1})^{r} \;=\; \binom{9}{r}\,2^{\,9-r}\,(-1)^{r}\,x^{\,2(9-r)}\,x^{-r}. $$ 合并 $x$ 的幂次：$x^{18-2r-r} = x^{18-3r}$。因此 $$ T_{r+1} = \binom{9}{r}\,2^{9-r}\,(-1)^{r}\,x^{18-3r}, \qquad r = 0, 1, \ldots, 9. $$

(b) $x^{6}$ 的系数 M1·A1

令指数等于 $6$：$18 - 3r = 6 \Rightarrow r = 4$。$r = 4$ 项为 $$ \binom{9}{4}\,2^{\,5}\,(-1)^{4}\,x^{6} \;=\; 126 \cdot 32 \cdot 1 \cdot x^{6} \;=\; 4032\,x^{6}. $$ 系数 $= \boxed{4032}$。

(c) 不含 $x$ 的项 M1·A1

令 $18 - 3r = 0 \Rightarrow r = 6$。$r = 6$ 项为 $$ \binom{9}{6}\,2^{\,3}\,(-1)^{6} \;=\; 84 \cdot 8 \cdot 1 \;=\; 672. $$ 所以展开式中的常数项（constant term）为 $\boxed{672}$。

常见陷阱：学生经常忽略第二项为负带来的 $(-1)^{r}$ 因子，或把 $x^{2}$ 的指数算错（应为 $2(9-r)$，而非 $(9-r)^{2}$）。一律先把两个 $x$ 的幂次合并为单一指数，再去找目标 $r$。

Q2MEDIUMPaper 1A1.10 Multiset Perms + Gap Method[7 marks]

MISSISSIPPI — (a) total distinguishable arrangements, (b) arrangements with no two I's adjacent.MISSISSIPPI —— (a) 可区分排列总数，(b) 任意两个 I 都不相邻的排列数。

Answers:答案： (a) $\dfrac{11!}{4!\,4!\,2!} = 34\,650$ · (b) $7350$

(a) Total arrangements M1·A1

Letter counts: $\mathrm{M}\!\times\!1,\ \mathrm{I}\!\times\!4,\ \mathrm{S}\!\times\!4,\ \mathrm{P}\!\times\!2$ — total $11$ letters. The number of distinguishable orderings of a multiset is $$ \frac{11!}{1!\,4!\,4!\,2!} \;=\; \frac{39\,916\,800}{1\cdot 24\cdot 24\cdot 2} \;=\; \frac{39\,916\,800}{1152} \;=\; 34\,650. $$ (The $1!$ for M is usually omitted; it equals $1$ and matches the prompt's expression.)

(b) No two I's adjacent — the gap method M1·M1·A1·R1·A1

Place the four I's last, so the adjacency constraint becomes a clean "choose distinct slots" problem.

Arrange the non-I letters first. These are seven letters: $\mathrm{M, S, S, S, S, P, P}$. Distinguishable orderings: $$ \frac{7!}{4!\,2!\,1!} \;=\; \frac{5040}{48} \;=\; 105. $$
Identify the gaps. Once the seven letters are placed in a row, there are exactly $\boxed{8}$ "slots" where an I could go without creating a new I-pair: one before the row, one after, and one between each consecutive pair of letters ($7$ letters $\Rightarrow 6$ between-slots $\Rightarrow 6 + 2 = 8$ slots total).
Drop four I's into four distinct slots. Since the I's are identical, this is a choice (not an arrangement) of $4$ slots from $8$: $$ \binom{8}{4} \;=\; 70. $$
Multiply. $$ 105 \times 70 \;=\; 7350. $$

So there are $\boxed{7350}$ arrangements with no two I's adjacent.

Why the gap method works. Treating the "no two adjacent" constraint as a slot-choice problem decouples the constraint from the arrangement of the other letters. If you instead try inclusion–exclusion on "at least one I-pair adjacent" it works, but the bookkeeping is brutal. The gap method is the canonical IB approach — identify the constrained letters and place them into gaps of the rest.

(a) 排列总数 M1·A1

字母个数：$\mathrm{M}\!\times\!1,\ \mathrm{I}\!\times\!4,\ \mathrm{S}\!\times\!4,\ \mathrm{P}\!\times\!2$ —— 共 $11$ 个字母。多重集（multiset）的可区分排列数为 $$ \frac{11!}{1!\,4!\,4!\,2!} \;=\; \frac{39\,916\,800}{1\cdot 24\cdot 24\cdot 2} \;=\; \frac{39\,916\,800}{1152} \;=\; 34\,650. $$ （M 对应的 $1!$ 通常省略，其值等于 $1$，与题目表达式一致。）

(b) 任意两个 I 都不相邻 —— 插空法（`gap method`） M1·M1·A1·R1·A1

把四个 I 放到最后再插，于是不相邻的约束就变成一个干净的"选不同槽位"问题。

先排其余字母。共七个字母：$\mathrm{M, S, S, S, S, P, P}$。可区分排列数： $$ \frac{7!}{4!\,2!\,1!} \;=\; \frac{5040}{48} \;=\; 105. $$
找出可插的"空隙"。把七个字母排成一行后，恰好有 $\boxed{8}$ 个"槽位"可以放 I 而不会形成新的 I-I 相邻：行首一个、行尾一个、每对相邻字母之间各一个（$7$ 个字母 $\Rightarrow 6$ 个中间槽 $\Rightarrow 6 + 2 = 8$ 个槽）。
从 $8$ 个槽中选 $4$ 个放 I。因为四个 I 是相同的，这是组合（choose）而非排列： $$ \binom{8}{4} \;=\; 70. $$
相乘。 $$ 105 \times 70 \;=\; 7350. $$

因此任意两个 I 都不相邻的排列数为 $\boxed{7350}$。

插空法为什么有效。把"任意两个不相邻"的约束转化为"选槽位"问题，可以让约束与其他字母的排列方式相互解耦。如果改用容斥原理（inclusion–exclusion）枚举"至少有一对 I 相邻"也能做，但记账非常痛苦。插空法是 IB 的标准做法 —— 识别出受约束的字母，然后把它们放入其他字母的空隙。

Q3HARDPaper 1A1.10 Consecutive Coefficient Ratios[7 marks]

Find $n$ and $r$ such that $\binom{n}{r} : \binom{n}{r+1} : \binom{n}{r+2} = 6 : 14 : 21$.求 $n$ 与 $r$，使 $\binom{n}{r} : \binom{n}{r+1} : \binom{n}{r+2} = 6 : 14 : 21$。

Answer:答案： $n = 9$, $r = 2$

(a) Ratio formula M1·A1

$$ \frac{\binom{n}{r+1}}{\binom{n}{r}} \;=\; \frac{\dfrac{n!}{(r+1)!\,(n-r-1)!}}{\dfrac{n!}{r!\,(n-r)!}} \;=\; \frac{r!\,(n-r)!}{(r+1)!\,(n-r-1)!} \;=\; \frac{n-r}{r+1}. $$ (Cancel $n!$ top and bottom, then cancel factorials: $r!/(r+1)! = 1/(r+1)$ and $(n-r)!/(n-r-1)! = n-r$.)

(b) Two equations, two unknowns M1·M1·A1·A1·A1

From the given ratio, $$ \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{14}{6} = \frac{7}{3} \quad \Longrightarrow \quad \frac{n-r}{r+1} = \frac{7}{3} \quad \Longrightarrow \quad 3(n-r) = 7(r+1) \quad \Longrightarrow \quad 3n = 10r + 7. \tag{1} $$ Apply the same ratio to the next pair (shift $r \mapsto r+1$ in part (a)): $$ \frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{21}{14} = \frac{3}{2} \quad \Longrightarrow \quad \frac{n-r-1}{r+2} = \frac{3}{2} \quad \Longrightarrow \quad 2(n-r-1) = 3(r+2) \quad \Longrightarrow \quad 2n = 5r + 8. \tag{2} $$ Eliminate $n$: multiply $(2)$ by $2$ to get $4n = 10r + 16$, then subtract $(1)$: $$ 4n - 3n \;=\; (10r + 16) - (10r + 7) \quad \Longrightarrow \quad n = 9. $$ Substitute back into $(2)$: $18 = 5r + 8 \Rightarrow r = 2$.

Verification

$\binom{9}{2} = 36$, $\binom{9}{3} = 84$, $\binom{9}{4} = 126$. Divide the triple by $\gcd = 6$: $36 : 84 : 126 = 6 : 14 : 21$. ✓

Why this question is “extending.” Most textbook problems stop at one ratio (i.e., a single equation $\binom{n}{r+1}/\binom{n}{r} = c$ and one of $n$ or $r$ already given). Using two consecutive ratios pins down both unknowns from a single chain — the key trick is recognising that the same lemma applies with $r \mapsto r+1$. This is a high-yield Paper 1A pattern; the same machinery handles $\binom{n}{r-1}:\binom{n}{r}:\binom{n}{r+1} = a:b:c$ problems.

(a) 比值公式 M1·A1

$$ \frac{\binom{n}{r+1}}{\binom{n}{r}} \;=\; \frac{\dfrac{n!}{(r+1)!\,(n-r-1)!}}{\dfrac{n!}{r!\,(n-r)!}} \;=\; \frac{r!\,(n-r)!}{(r+1)!\,(n-r-1)!} \;=\; \frac{n-r}{r+1}. $$ （上下消去 $n!$，再消阶乘：$r!/(r+1)! = 1/(r+1)$，$(n-r)!/(n-r-1)! = n-r$。）

(b) 两个未知数的两个方程 M1·M1·A1·A1·A1

由题目给定的比例， $$ \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{14}{6} = \frac{7}{3} \quad \Longrightarrow \quad \frac{n-r}{r+1} = \frac{7}{3} \quad \Longrightarrow \quad 3(n-r) = 7(r+1) \quad \Longrightarrow \quad 3n = 10r + 7. \tag{1} $$ 对下一对应用同样的比例（在 (a) 中把 $r$ 替换为 $r+1$）： $$ \frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{21}{14} = \frac{3}{2} \quad \Longrightarrow \quad \frac{n-r-1}{r+2} = \frac{3}{2} \quad \Longrightarrow \quad 2(n-r-1) = 3(r+2) \quad \Longrightarrow \quad 2n = 5r + 8. \tag{2} $$ 消去 $n$：将 $(2)$ 两边乘以 $2$ 得 $4n = 10r + 16$，再减去 $(1)$： $$ 4n - 3n \;=\; (10r + 16) - (10r + 7) \quad \Longrightarrow \quad n = 9. $$ 代回 $(2)$：$18 = 5r + 8 \Rightarrow r = 2$。

验证

$\binom{9}{2} = 36$、$\binom{9}{3} = 84$、$\binom{9}{4} = 126$。将三数除以 $\gcd = 6$：$36 : 84 : 126 = 6 : 14 : 21$。✓

本题的"延伸性"所在。多数教材题只给出一个比值（即单一方程 $\binom{n}{r+1}/\binom{n}{r} = c$，且 $n$ 或 $r$ 之一已给定）。利用两个相邻比值就能从同一条链上锁定两个未知数 —— 关键在于意识到把 $r$ 替换为 $r+1$ 时同一引理依旧成立。这是 Paper 1A 的高价值题型；同一套机制也能处理 $\binom{n}{r-1}:\binom{n}{r}:\binom{n}{r+1} = a:b:c$ 之类的问题。

Q4HARDPaper 1A1.9 Identity via Differentiation[8 marks]

Prove three weighted-sum identities for $\sum r \binom{n}{r}$, $\sum r(r-1)\binom{n}{r}$, and $\sum r^{2}\binom{n}{r}$ for $n \ge 2$.当 $n \ge 2$ 时，分别证明 $\sum r \binom{n}{r}$、$\sum r(r-1)\binom{n}{r}$ 与 $\sum r^{2}\binom{n}{r}$ 的三个加权求和恒等式。

All three identities proved; see derivation.三个恒等式均已证明；推导见下。

(a) First derivative M1·A1·R1

Start from the binomial expansion $$ (1+x)^{n} \;=\; \sum_{r=0}^{n}\binom{n}{r}\,x^{r}. \tag{$\star$} $$ Differentiate both sides with respect to $x$ (LHS via chain rule, RHS termwise): $$ n(1+x)^{n-1} \;=\; \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{r-1}. $$ (The $r=0$ term contributed $\binom{n}{0}$ — a constant — whose derivative is $0$, so the sum now starts at $r = 1$.) Substitute $x = 1$: $$ n\cdot 2^{\,n-1} \;=\; \sum_{r=1}^{n} r\,\binom{n}{r}. \qquad \blacksquare $$

(b) Second derivative M1·A1·R1

Differentiate the equation in (a) once more: $$ n(n-1)(1+x)^{n-2} \;=\; \sum_{r=2}^{n} r(r-1)\,\binom{n}{r}\,x^{r-2}. $$ The $r=1$ term contributed $r\binom{n}{r} = 1\cdot\binom{n}{1}$ — a constant — whose derivative is $0$, so the sum now starts at $r=2$. Substitute $x = 1$: $$ n(n-1)\cdot 2^{\,n-2} \;=\; \sum_{r=2}^{n} r(r-1)\,\binom{n}{r}. \qquad \blacksquare $$

(c) Combine via $r^{2} = r(r-1) + r$ M1·A1

Split the squared weight: $$ \sum_{r=0}^{n} r^{2}\binom{n}{r} \;=\; \sum_{r=0}^{n}\bigl[\,r(r-1) + r\,\bigr]\binom{n}{r} \;=\; \sum_{r=2}^{n} r(r-1)\binom{n}{r} \,+\, \sum_{r=1}^{n} r\binom{n}{r}. $$ (The $r=0$ and $r=1$ terms vanish in the $r(r-1)$ sum; the $r=0$ term vanishes in the $r$ sum.) Apply (a) and (b): $$ = n(n-1)\,2^{\,n-2} + n\,2^{\,n-1} \;=\; 2^{\,n-2}\bigl[\,n(n-1) + 2n\,\bigr] \;=\; 2^{\,n-2}\cdot n(n+1) \;=\; n(n+1)\,2^{\,n-2}. \qquad \blacksquare $$

The pattern. Differentiating $(1+x)^{n}$ $k$ times and substituting $x = 1$ gives a closed form for $\sum r(r-1)\cdots(r-k+1)\,\binom{n}{r}$ — these are the falling-factorial weights. Any polynomial weight $p(r)$ can be expanded as a sum of falling factorials and evaluated in one line. This is the bridge from binomial sums to (HL) discrete moment problems — the same algebra computes $E(X)$ and $E(X^{2})$ for $X \sim \mathrm{Bin}(n, \tfrac{1}{2})$.

(a) 一阶导数 M1·A1·R1

从二项展开式出发： $$ (1+x)^{n} \;=\; \sum_{r=0}^{n}\binom{n}{r}\,x^{r}. \tag{$\star$} $$ 两边对 $x$ 求导（左边用链式法则，右边逐项求导）： $$ n(1+x)^{n-1} \;=\; \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{r-1}. $$ （$r=0$ 项 $\binom{n}{0}$ 是常数，导数为 $0$，所以求和从 $r = 1$ 开始。）代入 $x = 1$： $$ n\cdot 2^{\,n-1} \;=\; \sum_{r=1}^{n} r\,\binom{n}{r}. \qquad \blacksquare $$

(b) 二阶导数 M1·A1·R1

对 (a) 中等式再求一次导： $$ n(n-1)(1+x)^{n-2} \;=\; \sum_{r=2}^{n} r(r-1)\,\binom{n}{r}\,x^{r-2}. $$ $r=1$ 项 $r\binom{n}{r} = 1\cdot\binom{n}{1}$ 是常数，导数为 $0$，故求和从 $r=2$ 开始。代入 $x = 1$： $$ n(n-1)\cdot 2^{\,n-2} \;=\; \sum_{r=2}^{n} r(r-1)\,\binom{n}{r}. \qquad \blacksquare $$

(c) 利用 $r^{2} = r(r-1) + r$ 合并 M1·A1

将平方权重拆分： $$ \sum_{r=0}^{n} r^{2}\binom{n}{r} \;=\; \sum_{r=0}^{n}\bigl[\,r(r-1) + r\,\bigr]\binom{n}{r} \;=\; \sum_{r=2}^{n} r(r-1)\binom{n}{r} \,+\, \sum_{r=1}^{n} r\binom{n}{r}. $$ （$r(r-1)$ 求和中 $r=0$、$r=1$ 项消失；$r$ 求和中 $r=0$ 项消失。）应用 (a) 与 (b)： $$ = n(n-1)\,2^{\,n-2} + n\,2^{\,n-1} \;=\; 2^{\,n-2}\bigl[\,n(n-1) + 2n\,\bigr] \;=\; 2^{\,n-2}\cdot n(n+1) \;=\; n(n+1)\,2^{\,n-2}. \qquad \blacksquare $$

规律。对 $(1+x)^{n}$ 求 $k$ 次导后代入 $x = 1$，可得 $\sum r(r-1)\cdots(r-k+1)\,\binom{n}{r}$ 的闭合形式 —— 这就是降阶乘（falling factorial）权重。任意多项式权重 $p(r)$ 都可以展开为若干降阶乘之和，从而一行算出。这正是二项求和通向 HL 离散矩问题的桥梁 —— 同一套代数也能算出 $X \sim \mathrm{Bin}(n, \tfrac{1}{2})$ 的 $E(X)$ 与 $E(X^{2})$。

Section B — Worked SolutionsB 节 —— 详细解析

Q5HARDPaper 1B1.9 Binomial — System & Largest Coefficient[12 marks]

$(1 + ax)^{n}$ has coefficient of $x$ equal to $24$ and coefficient of $x^{2}$ equal to $252$. Find $n$, $a$, the coefficient of $x^{3}$, and the largest coefficient in the expansion.$(1 + ax)^{n}$ 中 $x$ 的系数为 $24$，$x^{2}$ 的系数为 $252$。求 $n$、$a$、$x^{3}$ 的系数，以及展开式中的最大系数。

Answers:答案： (a) $T_{r+1} = \binom{n}{r}a^{r}x^{r}$ · (c) $n = 8,\ a = 3$ · (d) $1512$ · (e) $T_{7}$ is largest with coefficient $20\,412$ (at $r = 6$).$T_{7}$ 最大，系数为 $20\,412$（$r = 6$）。

(a) General term A1

With $a$-of-binomial $= 1$ and $b$-of-binomial $= ax$: $$ T_{r+1} \;=\; \binom{n}{r}\,(1)^{n-r}\,(ax)^{r} \;=\; \binom{n}{r}\,a^{r}\,x^{r}. $$

(b) Coefficient of $x^{k}$ M1·A1

$T_{r+1} = \binom{n}{r}a^{r}x^{r}$, so $x^{k}$ comes from $r = k$ and its coefficient is $\binom{n}{k}\,a^{k}$. $\blacksquare$

(c)(i) Two equations M1·A1

$$ \text{coef of }x: \;\; \binom{n}{1}\,a = na = 24. \qquad\quad \text{coef of }x^{2}: \;\; \binom{n}{2}\,a^{2} = \frac{n(n-1)}{2}\,a^{2} = 252. $$

(c)(ii) Solve M1·A1·A1

From $na = 24$: $a = 24/n$. Substitute into $\tfrac{n(n-1)}{2}a^{2} = 252$: $$ \frac{n(n-1)}{2}\cdot\frac{576}{n^{2}} \;=\; 252 \quad \Longrightarrow \quad \frac{288(n-1)}{n} \;=\; 252 \quad \Longrightarrow \quad \frac{n-1}{n} = \frac{252}{288} = \frac{7}{8}. $$ Cross-multiply: $8(n-1) = 7n \Rightarrow n = 8$. Then $a = 24/8 = 3$. $\;\Box$

(d) Coefficient of $x^{3}$ A1

By (b): $\binom{8}{3}\,3^{3} = 56 \cdot 27 = 1512$.

(e) Largest coefficient M1·A1·A1

The coefficients are $c_{r} = \binom{8}{r}\,3^{r}$ for $r = 0, 1, \ldots, 8$. Form the ratio of consecutive coefficients: $$ \frac{c_{r+1}}{c_{r}} \;=\; \frac{\binom{8}{r+1}\,3^{r+1}}{\binom{8}{r}\,3^{r}} \;=\; 3\cdot\frac{8-r}{r+1}. $$ Coefficients increase while this ratio $> 1$: $$ 3(8 - r) > r + 1 \;\Longleftrightarrow\; 24 - 3r > r + 1 \;\Longleftrightarrow\; 4r < 23 \;\Longleftrightarrow\; r < 5.75. $$ So $c_{r+1} > c_{r}$ for $r = 0, 1, \ldots, 5$, and $c_{r+1} < c_{r}$ for $r \ge 6$. The maximum is therefore at $r = 6$ — this is $T_{7}$ in the expansion. Its coefficient is $$ c_{6} \;=\; \binom{8}{6}\,3^{6} \;=\; 28\cdot 729 \;=\; 20\,412. $$

Sanity-check the table. $c_{r} = \binom{8}{r}3^{r}$ gives $1,\,24,\,252,\,1512,\,5670,\,13\,608,\,20\,412,\,17\,496,\,6561$ — increasing then decreasing, peaking at $r = 6$. This consecutive-ratio test is the canonical IB approach to "largest coefficient" questions; it works for any $(1 + bx)^{n}$ and avoids brute-force enumeration.

(a) 通项 A1

取二项中 $a$-端 $= 1$、$b$-端 $= ax$： $$ T_{r+1} \;=\; \binom{n}{r}\,(1)^{n-r}\,(ax)^{r} \;=\; \binom{n}{r}\,a^{r}\,x^{r}. $$

(b) $x^{k}$ 的系数 M1·A1

$T_{r+1} = \binom{n}{r}a^{r}x^{r}$，所以 $x^{k}$ 来自 $r = k$ 项，其系数为 $\binom{n}{k}\,a^{k}$。$\blacksquare$

(c)(i) 两个方程 M1·A1

$x$ 的系数：$\binom{n}{1}\,a = na = 24$。$x^{2}$ 的系数：$\binom{n}{2}\,a^{2} = \tfrac{n(n-1)}{2}\,a^{2} = 252$。即 $$ na = 24, \qquad \tfrac{n(n-1)}{2}\,a^{2} = 252. $$

(c)(ii) 求解 M1·A1·A1

由 $na = 24$：$a = 24/n$。代入 $\tfrac{n(n-1)}{2}a^{2} = 252$： $$ \frac{n(n-1)}{2}\cdot\frac{576}{n^{2}} \;=\; 252 \quad \Longrightarrow \quad \frac{288(n-1)}{n} \;=\; 252 \quad \Longrightarrow \quad \frac{n-1}{n} = \frac{252}{288} = \frac{7}{8}. $$ 交叉相乘：$8(n-1) = 7n \Rightarrow n = 8$。再得 $a = 24/8 = 3$。$\;\Box$

(d) $x^{3}$ 的系数 A1

由 (b)：$\binom{8}{3}\,3^{3} = 56 \cdot 27 = 1512$。

(e) 最大系数 M1·A1·A1

系数为 $c_{r} = \binom{8}{r}\,3^{r}$，$r = 0, 1, \ldots, 8$。考虑相邻系数之比： $$ \frac{c_{r+1}}{c_{r}} \;=\; \frac{\binom{8}{r+1}\,3^{r+1}}{\binom{8}{r}\,3^{r}} \;=\; 3\cdot\frac{8-r}{r+1}. $$ 系数递增当且仅当该比值 $> 1$： $$ 3(8 - r) > r + 1 \;\Longleftrightarrow\; 24 - 3r > r + 1 \;\Longleftrightarrow\; 4r < 23 \;\Longleftrightarrow\; r < 5.75. $$ 所以 $r = 0, 1, \ldots, 5$ 时 $c_{r+1} > c_{r}$，$r \ge 6$ 时 $c_{r+1} < c_{r}$。最大值出现在 $r = 6$ —— 即展开式中的 $T_{7}$。其系数为 $$ c_{6} \;=\; \binom{8}{6}\,3^{6} \;=\; 28\cdot 729 \;=\; 20\,412. $$

用系数表反查。$c_{r} = \binom{8}{r}3^{r}$ 依次为 $1,\,24,\,252,\,1512,\,5670,\,13\,608,\,20\,412,\,17\,496,\,6561$ —— 先递增后递减，在 $r = 6$ 取峰值。这种相邻比值法（consecutive-ratio test）是 IB"求最大系数"问题的标准做法；适用于任意 $(1 + bx)^{n}$，避免暴力枚举。

Q6HARDPaper 1B1.10 Hockey-Stick Identity[13 marks]

$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$. (a) Verify $r=2, n=5$. (b) Combinatorial proof. (c) Evaluate $\sum_{k=3}^{20}\binom{k}{3}$. (d) Prove $1+2+\cdots+n = n(n+1)/2$.$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$。(a) 验证 $r=2, n=5$。(b) 组合学证明。(c) 计算 $\sum_{k=3}^{20}\binom{k}{3}$。(d) 证明 $1+2+\cdots+n = n(n+1)/2$。

Answers:答案： (a) Both sides $= 20$两边均等于 $20$ · (c) $\binom{21}{4} = 5985$ · (b) and (d) proofs below.(b) 与 (d) 的证明见下。

(a) Verification, $r = 2, n = 5$ M1·A1·A1

LHS:

$$ \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} \;=\; 1 + 3 + 6 + 10 \;=\; 20. $$ RHS: $\binom{6}{3} = \dfrac{6!}{3!\,3!} = \dfrac{6\cdot 5\cdot 4}{6} = 20$. LHS $=$ RHS. ✓

(b) Combinatorial proof — classify by largest element M1·M1·R1·R1·A1

Common count. Let $S = \{1, 2, 3, \ldots, n+1\}$. We count the $(r+1)$-element subsets of $S$ in two different ways.

Right-hand side. Directly: the number of $(r+1)$-subsets of an $(n+1)$-set is $\binom{n+1}{r+1}$.

Left-hand side. Classify each $(r+1)$-subset by its largest element. Suppose the largest element is $k$. Since the subset has $r+1$ elements and the largest is $k$, the smallest possible $k$ is $r+1$ (subset $= \{1, 2, \ldots, r+1\}$) and the largest possible $k$ is $n+1$. For each $k \in \{r+1, r+2, \ldots, n+1\}$:

The element $k$ is in the subset.
The remaining $r$ elements must be chosen from $\{1, 2, \ldots, k-1\}$ — that's $\binom{k-1}{r}$ choices.

Therefore the total number of $(r+1)$-subsets is

$$ \sum_{k=r+1}^{n+1}\binom{k-1}{r} \;=\; \sum_{j = r}^{n}\binom{j}{r} \quad\text{(re-index $j = k-1$)} \;=\; \binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r}. $$

Two counts of the same set are equal:

$$ \binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} \;=\; \binom{n+1}{r+1}. \qquad \blacksquare $$

(c) Application M1·A1

With $r = 3$ and $n = 20$ in the identity: $$ \sum_{k=3}^{20}\binom{k}{3} \;=\; \binom{21}{4} \;=\; \frac{21\cdot 20\cdot 19\cdot 18}{4!} \;=\; \frac{143\,640}{24} \;=\; 5985. $$

(d) Triangular-number derivation M1·M1·A1

Note $k = \binom{k}{1}$ (choosing $1$ from $k$). Apply the identity with $r = 1$: $$ \binom{1}{1} + \binom{2}{1} + \binom{3}{1} + \cdots + \binom{n}{1} \;=\; \binom{n+1}{2}. $$ Substitute $\binom{k}{1} = k$ and $\binom{n+1}{2} = \dfrac{(n+1)n}{2}$: $$ 1 + 2 + 3 + \cdots + n \;=\; \frac{n(n+1)}{2}. \qquad \blacksquare $$

Why the hockey stick is named the hockey stick. Highlight the entries on Pascal's triangle: each summand $\binom{k}{r}$ runs diagonally down one side, and the answer $\binom{n+1}{r+1}$ sits one row below at the elbow — the highlighted shape looks like a hockey stick. The combinatorial proof above generalises every identity of the form "sum down a diagonal of Pascal's triangle equals one entry below the elbow" — including the formulas for higher-order figurate numbers ($\binom{k}{2}$ gives triangular numbers, $\binom{k}{3}$ gives tetrahedral numbers, etc.).

(a) 验证 $r = 2, n = 5$ M1·A1·A1

LHS：

$$ \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} \;=\; 1 + 3 + 6 + 10 \;=\; 20. $$ RHS：$\binom{6}{3} = \dfrac{6!}{3!\,3!} = \dfrac{6\cdot 5\cdot 4}{6} = 20$。两边相等。✓

(b) 组合学证明 —— 按最大元素分类 M1·M1·R1·R1·A1

共同计数对象。设 $S = \{1, 2, 3, \ldots, n+1\}$。我们用两种不同方法计数 $S$ 的所有 $(r+1)$-元子集。

右边。直接计数：$(n+1)$-元集合的 $(r+1)$-元子集数为 $\binom{n+1}{r+1}$。

左边。按每个 $(r+1)$-子集的最大元素进行分类。设最大元素为 $k$。由于子集含 $r+1$ 个元素且最大为 $k$，$k$ 最小可取 $r+1$（对应子集 $\{1, 2, \ldots, r+1\}$），最大可取 $n+1$。对每个 $k \in \{r+1, r+2, \ldots, n+1\}$：

$k$ 必在该子集中。
余下 $r$ 个元素须从 $\{1, 2, \ldots, k-1\}$ 中选 —— 共 $\binom{k-1}{r}$ 种选法。

因此所有 $(r+1)$-子集的总数为

（令 $j = k - 1$ 换元：） $$ \sum_{k=r+1}^{n+1}\binom{k-1}{r} \;=\; \sum_{j = r}^{n}\binom{j}{r} \;=\; \binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r}. $$

同一集合的两种计数必然相等：

$$ \binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} \;=\; \binom{n+1}{r+1}. \qquad \blacksquare $$

(c) 应用 M1·A1

在恒等式中取 $r = 3$、$n = 20$： $$ \sum_{k=3}^{20}\binom{k}{3} \;=\; \binom{21}{4} \;=\; \frac{21\cdot 20\cdot 19\cdot 18}{4!} \;=\; \frac{143\,640}{24} \;=\; 5985. $$

(d) 三角形数公式的推导 M1·M1·A1

注意到 $k = \binom{k}{1}$（从 $k$ 个中选 $1$ 个）。在恒等式中取 $r = 1$： $$ \binom{1}{1} + \binom{2}{1} + \binom{3}{1} + \cdots + \binom{n}{1} \;=\; \binom{n+1}{2}. $$ 代入 $\binom{k}{1} = k$ 与 $\binom{n+1}{2} = \dfrac{(n+1)n}{2}$： $$ 1 + 2 + 3 + \cdots + n \;=\; \frac{n(n+1)}{2}. \qquad \blacksquare $$

"曲棍球棒"得名的由来。在 Pascal 三角形上高亮各项：每个加项 $\binom{k}{r}$ 沿对角线向下延伸，而答案 $\binom{n+1}{r+1}$ 位于"弯钩处"下一行 —— 高亮出来的形状像极了一根曲棍球棒。上述组合学证明可推广到所有形如"沿 Pascal 三角形对角线求和等于弯钩处下方某一格"的恒等式 —— 也包括高阶 figurate 数（图形数）的公式（$\binom{k}{2}$ 给出三角形数，$\binom{k}{3}$ 给出四面体数等等）。

Q11HARDPaper 1B1.10 Lattice Paths — Combinations on a Grid[11 marks]

Monotone lattice paths from $(0,0)$ to $(5,4)$ on the integer grid (each step is $R$ or $U$). Count: total, through $(3,2)$, not through $(3,2)$, through $(3,2)$ but not $(1,1)$, and the probability that a random path passes through both $(1,1)$ and $(3,2)$.整数格上从 $(0,0)$ 到 $(5,4)$ 的单调格点路径（每步为 $R$ 或 $U$）。求：总数、经过 $(3,2)$ 的、不经过 $(3,2)$ 的、经过 $(3,2)$ 但不经过 $(1,1)$ 的，以及随机路径同时经过 $(1,1)$ 与 $(3,2)$ 的概率。

Answers:答案： (a) $\binom{9}{4} = 126$ · (b) $60$ · (c) $66$ · (d) $24$ · (e) $\dfrac{36}{126} = \dfrac{2}{7}$

(a) Total monotone paths M1·A1

Each path from $(0,0)$ to $(5,4)$ takes exactly $5$ right-steps and $4$ up-steps, in some order: a sequence of $9$ symbols, of which $5$ are $R$ and $4$ are $U$. The number of such sequences is the number of ways to choose which $4$ of the $9$ positions are $U$'s (equivalently which $5$ are $R$'s): $$ \binom{9}{4} \;=\; \binom{9}{5} \;=\; \frac{9!}{4!\,5!} \;=\; \frac{9\cdot 8\cdot 7\cdot 6}{4!} \;=\; \frac{3024}{24} \;=\; 126. $$

(b) Paths through $(3,2)$ M1·M1·A1

A path passes through $(3,2)$ iff it can be split into two independent sub-paths: $(0,0)\to(3,2)$ followed by $(3,2)\to(5,4)$. Count each leg:

$(0,0)\to(3,2)$: $3$ rights $+$ $2$ ups in any order $\Rightarrow \binom{5}{2} = 10$ paths.
$(3,2)\to(5,4)$: $2$ rights $+$ $2$ ups in any order $\Rightarrow \binom{4}{2} = 6$ paths.

By the multiplication principle, total $= 10\cdot 6 = 60$.

(c) Paths NOT through $(3,2)$ A1

Complement of (b): $126 - 60 = 66$.

(d) Through $(3,2)$ but NOT through $(1,1)$ M1·M1·A1·A1

Use inclusion–exclusion on the set of paths through $(3,2)$: $$ \#\bigl(\text{thru }(3,2)\bigr) \;=\; \#\bigl(\text{thru }(1,1)\text{ AND }(3,2)\bigr) \;+\; \#\bigl(\text{thru }(3,2)\text{ but NOT }(1,1)\bigr). $$ Count paths through both $(1,1)$ and $(3,2)$ — split into three legs:

$(0,0)\to(1,1)$: $1R + 1U \Rightarrow \binom{2}{1} = 2$.
$(1,1)\to(3,2)$: $2R + 1U \Rightarrow \binom{3}{1} = 3$.
$(3,2)\to(5,4)$: $2R + 2U \Rightarrow \binom{4}{2} = 6$.

Product: $2\cdot 3\cdot 6 = 36$. Then $\#\bigl(\text{thru }(3,2)\text{, not }(1,1)\bigr) = 60 - 36 = 24$.

(e) Probability of passing through both checkpoints R1·A1

Uniform random path from (a): each of the $126$ paths is equally likely. From (d)'s subcount, $36$ pass through both: $$ P\bigl(\text{thru }(1,1)\text{ and }(3,2)\bigr) \;=\; \frac{36}{126} \;=\; \frac{2}{7}. $$

Multiplication principle, by sub-segments. Lattice-path counting is the canonical visual setting for combinations: each path corresponds to a binary word of $R$s and $U$s, and $\binom{m+n}{m}$ counts them. The technique generalises directly to constraint counting — "must pass through $P_{1}, P_{2}, \ldots, P_{k}$" splits the path into $k+1$ independent legs whose counts multiply. "Must avoid $P$" is the complement; "avoid one but include another" is inclusion–exclusion. This same machinery underlies Catalan-number counts (paths that don't cross a diagonal), which is the next extending question if you've enjoyed this one.

(a) 单调路径总数 M1·A1

从 $(0,0)$ 到 $(5,4)$ 的每条路径恰由 $5$ 步右移和 $4$ 步上移按某顺序构成：长度为 $9$ 的符号序列，其中 $5$ 个为 $R$、$4$ 个为 $U$。这样的序列数等于在 $9$ 个位置中选哪 $4$ 个放 $U$（等价地选 $5$ 个放 $R$）： $$ \binom{9}{4} \;=\; \binom{9}{5} \;=\; \frac{9!}{4!\,5!} \;=\; \frac{9\cdot 8\cdot 7\cdot 6}{4!} \;=\; \frac{3024}{24} \;=\; 126. $$

(b) 经过 $(3,2)$ 的路径数 M1·M1·A1

路径经过 $(3,2)$ 当且仅当其可拆分为两段独立子路径：$(0,0)\to(3,2)$ 与 $(3,2)\to(5,4)$。分别计数：

$(0,0)\to(3,2)$：$3$ 步右 + $2$ 步上，任意顺序 $\Rightarrow \binom{5}{2} = 10$ 条。
$(3,2)\to(5,4)$：$2$ 步右 + $2$ 步上，任意顺序 $\Rightarrow \binom{4}{2} = 6$ 条。

由乘法原理（multiplication principle），总数 $= 10\cdot 6 = 60$。

(c) 不经过 $(3,2)$ 的路径数 A1

取 (b) 的补集：$126 - 60 = 66$。

(d) 经过 $(3,2)$ 但不经过 $(1,1)$ M1·M1·A1·A1

设 $A$ = 经过 $(1,1)$ 的路径集合，$B$ = 经过 $(3,2)$ 的路径集合。对 $B$ 做容斥（inclusion–exclusion）： $$ \#(B) \;=\; \#(A \cap B) \;+\; \#(B \setminus A). $$ 计数同时经过 $(1,1)$ 与 $(3,2)$ 的路径 —— 拆为三段：

$(0,0)\to(1,1)$：$1R + 1U \Rightarrow \binom{2}{1} = 2$。
$(1,1)\to(3,2)$：$2R + 1U \Rightarrow \binom{3}{1} = 3$。
$(3,2)\to(5,4)$：$2R + 2U \Rightarrow \binom{4}{2} = 6$。

乘积：$2\cdot 3\cdot 6 = 36$。再得 $\#(B \setminus A) = 60 - 36 = 24$（即经过 $(3,2)$ 但不经过 $(1,1)$ 的路径数）。

(e) 同时经过两个检查点的概率 R1·A1

(a) 中均匀随机抽取路径：$126$ 条路径每条等概率。由 (d) 的子计数知有 $36$ 条同时经过两点： $$ P(A \cap B) \;=\; \frac{36}{126} \;=\; \frac{2}{7}. $$ （$A \cap B$ = 同时经过 $(1,1)$ 与 $(3,2)$。）

按子段使用乘法原理。格点路径计数是组合数的经典视觉化设定：每条路径对应一个由 $R$ 和 $U$ 组成的二进制串，$\binom{m+n}{m}$ 即为其数目。这一技巧可直接推广到约束计数 —— "必须经过 $P_{1}, P_{2}, \ldots, P_{k}$" 将路径拆为 $k+1$ 段独立子路径，路径数相乘即可。"必须绕开 $P$"取补集；"绕开一个但必经另一个"用容斥。同样这套机制也支撑 Catalan 数（不越过某条对角线的路径数），若本题做得顺手，下一道延伸题正是该方向。

Paper 2 — Worked Solutions第二卷 —— 详细解析

Q7MEDIUMPaper 21.10 Combinations + Conditional Probability[7 marks]

Committee of $4$ from $5$ IB $+$ $7$ AP students. (a) P(exactly $2$ IB), (b) P(at least $2$ IB), (c) P(exactly $2$ IB $|$ at least $1$ IB).从 $5$ 名 IB + $7$ 名 AP 学生中选 $4$ 人组成委员会。(a) P(恰好 $2$ 名 IB)，(b) P(至少 $2$ 名 IB)，(c) P(恰好 $2$ 名 IB $|$ 至少 $1$ 名 IB)。

Answers:答案： (a) $\dfrac{14}{33}$ · (b) $\dfrac{19}{33}$ · (c) $\dfrac{21}{46}$

Set-up

Sample space size $= \binom{12}{4} = 495$. Each of the $495$ committees is equally likely.

(a) Exactly $2$ IB M1·A1

Choose $2$ of $5$ IB and $2$ of $7$ AP: $$ P(\text{2 IB}) \;=\; \frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}} \;=\; \frac{10\cdot 21}{495} \;=\; \frac{210}{495} \;=\; \frac{14}{33}. $$

(b) At least $2$ IB — via the complement M1·M1·A1

$P(\ge 2) = 1 - P(0) - P(1)$. $$ P(0) \;=\; \frac{\binom{5}{0}\binom{7}{4}}{495} \;=\; \frac{35}{495} \;=\; \frac{7}{99}, \qquad P(1) \;=\; \frac{\binom{5}{1}\binom{7}{3}}{495} \;=\; \frac{5\cdot 35}{495} \;=\; \frac{175}{495} \;=\; \frac{35}{99}. $$ Therefore $$ P(\ge 2) \;=\; 1 - \frac{7}{99} - \frac{35}{99} \;=\; \frac{99 - 7 - 35}{99} \;=\; \frac{57}{99} \;=\; \frac{19}{33}. $$

(c) Conditional probability M1·A1

$\{\text{exactly 2 IB}\} \subseteq \{\text{at least 1 IB}\}$, so $$ P(\text{exactly 2} \mid \text{at least 1}) \;=\; \frac{P(\text{exactly 2})}{P(\text{at least 1})} \;=\; \frac{14/33}{1 - 7/99} \;=\; \frac{14/33}{92/99}. $$ Multiply numerator and denominator by $99$: $= \dfrac{14\cdot 3}{92} = \dfrac{42}{92} = \dfrac{21}{46}.$

Tip. In (b), always check that "at least $k$" is small enough to compute the complement cheaply — here $P(0)$ and $P(1)$ are two terms vs. three for $P(2) + P(3) + P(4)$. Method marks are awarded for the strategy, so state “use complement” explicitly.

列式

样本空间大小 $= \binom{12}{4} = 495$。每个委员会等概率出现。

(a) 恰好 $2$ 名 IB M1·A1

从 $5$ 名 IB 中选 $2$、从 $7$ 名 AP 中选 $2$： $$ P(\text{2 IB}) \;=\; \frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}} \;=\; \frac{10\cdot 21}{495} \;=\; \frac{210}{495} \;=\; \frac{14}{33}. $$

(b) 至少 $2$ 名 IB —— 用补集 M1·M1·A1

$P(\ge 2) = 1 - P(0) - P(1)$。 $$ P(0) \;=\; \frac{\binom{5}{0}\binom{7}{4}}{495} \;=\; \frac{35}{495} \;=\; \frac{7}{99}, \qquad P(1) \;=\; \frac{\binom{5}{1}\binom{7}{3}}{495} \;=\; \frac{5\cdot 35}{495} \;=\; \frac{175}{495} \;=\; \frac{35}{99}. $$ 因此 $$ P(\ge 2) \;=\; 1 - \frac{7}{99} - \frac{35}{99} \;=\; \frac{99 - 7 - 35}{99} \;=\; \frac{57}{99} \;=\; \frac{19}{33}. $$

(c) 条件概率 M1·A1

记 $X$ = 抽中的 IB 学生数。事件 $\{X = 2\} \subseteq \{X \ge 1\}$，所以 $$ P(X = 2 \mid X \ge 1) \;=\; \frac{P(X = 2)}{P(X \ge 1)} \;=\; \frac{14/33}{1 - 7/99} \;=\; \frac{14/33}{92/99}. $$ 分子分母同乘 $99$：$= \dfrac{14\cdot 3}{92} = \dfrac{42}{92} = \dfrac{21}{46}$。

小贴士。做 (b) 时，先检查"至少 $k$"是否便于用补集 —— 这里 $P(0) + P(1)$ 是两项，而正面计算 $P(2) + P(3) + P(4)$ 是三项。方法分认的是策略，所以要明确写出"使用补集"。

Q8MEDIUMPaper 21.10 Linear Perms — Gap Method[7 marks]

$6$ boys and $4$ girls in a row of $10$. (a) Total. (b) No two girls adjacent. (c) Probability. (d) $B_{1}\,G\,B_{2}$ or $B_{2}\,G\,B_{1}$ as a consecutive block.$6$ 男 $4$ 女坐成 $10$ 人一行。(a) 总数。(b) 任意两女不相邻。(c) 概率。(d) $B_{1}\,G\,B_{2}$ 或 $B_{2}\,G\,B_{1}$ 作为连续块。

Answers:答案： (a) $10! = 3\,628\,800$ · (b) $604\,800$ · (c) $\dfrac{1}{6}$ · (d) $80\,640$

(a) Total A1

All $10$ students distinguishable, all seats distinguishable: $10! = 3\,628\,800$.

(b) No two girls adjacent — gap method (linear version) M1·M1·A1

Arrange the boys first: $6! = 720$ ways. They create $7$ gaps (one before the row, $5$ between consecutive boys, one after). Place the $4$ girls into $4$ distinct gaps and order them: $$ \binom{7}{4}\cdot 4! \;=\; 35\cdot 24 \;=\; 840. $$ Total: $720 \cdot 840 = 604\,800$.

(c) Probability A1

$$ P(\text{no two girls adjacent}) \;=\; \frac{604\,800}{3\,628\,800} \;=\; \frac{1}{6}. $$

(d) Specific triple $B_{1}\,G\,B_{2}$ as a block M1·A1

Treat the three students as a single block. The block has internal arrangements $B_{1}\,G\,B_{2}$ or $B_{2}\,G\,B_{1}$ (the prompt fixes $G$ in the middle), so $2$ internal orderings. With the block + the remaining $7$ students $= 8$ units to arrange linearly: $$ 2 \cdot 8! \;=\; 2 \cdot 40\,320 \;=\; 80\,640. $$

"Gap method" vs. "block method". Part (b) uses the gap method (constrained items go into gaps of the rest). Part (d) uses the block method (constrained items glue together into a single super-unit). Knowing which to reach for is the IB skill being tested — in general, separation constraints want gaps, togetherness constraints want blocks.

(a) 总数 A1

$10$ 名学生彼此可区分，$10$ 张椅子彼此可区分：$10! = 3\,628\,800$。

(b) 任意两女不相邻 —— 插空法（直线版） M1·M1·A1

先排男生：$6! = 720$ 种。$6$ 名男生产生 $7$ 个空隙（行首一个、相邻男生之间 $5$ 个、行尾一个）。从 $7$ 个空隙中选 $4$ 个安置女生，再排女生： $$ \binom{7}{4}\cdot 4! \;=\; 35\cdot 24 \;=\; 840. $$ 总数：$720 \cdot 840 = 604\,800$。

(c) 概率 A1

（记事件 $N$ = 任意两女不相邻。） $$ P(N) \;=\; \frac{604\,800}{3\,628\,800} \;=\; \frac{1}{6}. $$

(d) 指定三人 $B_{1}\,G\,B_{2}$ 作为块 M1·A1

将三人视为一个块。块内排法为 $B_{1}\,G\,B_{2}$ 或 $B_{2}\,G\,B_{1}$（题目指定 $G$ 在中间），即 $2$ 种内部顺序。块加上其余 $7$ 人 $= 8$ 个单位排成一行： $$ 2 \cdot 8! \;=\; 2 \cdot 40\,320 \;=\; 80\,640. $$

"插空法" vs."捆绑法"。(b) 用的是插空法（受约束的元素放入其余元素的空隙）。(d) 用的是捆绑法（block method，受约束的元素粘成一个超级单位）。判断该用哪一种正是 IB 在考的技能 —— 总体上，分离类约束用插空，聚合类约束用捆绑。

Q9HARDPaper 21.10 Circular Perms — Cases & Gaps[7 marks]

$10$ people incl. sisters A, B, C around a round table. (a) Total. (b) Sisters all together. (c) No two of A, B, C adjacent.$10$ 人（含三姐妹 A、B、C）围圆桌。(a) 总数。(b) 三姐妹全相邻。(c) A、B、C 任意两人都不相邻。

Answers:答案： (a) $9! = 362\,880$ · (b) $30\,240$ · (c) $151\,200$

(a) Total circular arrangements A1

$n$ distinguishable people round a circular table: $(n-1)!$ arrangements. With $n = 10$: $9! = 362\,880$.

(b) Sisters all together — block method on a circle M1·A1

Treat A, B, C as a block of $3$. Internal orderings: $3! = 6$. With the block + the remaining $7$ people, we have $8$ units to seat round the circle: $(8 - 1)! = 7! = 5040$ arrangements. Total: $$ 3!\cdot 7! \;=\; 6\cdot 5040 \;=\; 30\,240. $$

(c) No two sisters adjacent — gap method on a circle M1·M1·A1·A1

Seat the $7$ non-sister people round the table. Circular arrangement: $(7-1)! = 6! = 720$ ways.
Identify the gaps. $7$ people around a circle create exactly $7$ gaps between consecutive people. (Note: a circle of $n$ people has $n$ gaps, not $n+1$ like a row.)
Place A, B, C into $3$ distinct gaps and order them. $\binom{7}{3}\cdot 3! = 35\cdot 6 = 210$ ways.
Multiply. $720 \cdot 210 = 151\,200$.

Circle gaps vs. line gaps. The most common slip on this question is reusing the "row" gap count (which would give $8$ gaps from $7$ items). On a circle, $n$ items give exactly $n$ gaps because the row "wraps around" — the slot “after the last person” and “before the first person” are the same slot. Always draw a small circle of $4$–$5$ dots to convince yourself before counting.

(a) 圆排列总数 A1

$n$ 个可区分的人围圆桌坐下：圆排列（circular permutation）数为 $(n-1)!$。$n = 10$ 时：$9! = 362\,880$。

(b) 三姐妹全相邻 —— 圆上的捆绑法 M1·A1

把 A、B、C 视为一个 $3$ 人块。块内顺序：$3! = 6$。块 + 其余 $7$ 人 $= 8$ 个单位围圆排列：$(8 - 1)! = 7! = 5040$。总数： $$ 3!\cdot 7! \;=\; 6\cdot 5040 \;=\; 30\,240. $$

(c) 三姐妹两两不相邻 —— 圆上的插空法 M1·M1·A1·A1

先让其余 $7$ 名非姐妹围桌就座。圆排列：$(7-1)! = 6! = 720$ 种。
找空隙。$7$ 个人围一圈，相邻人之间恰有 $7$ 个空隙。（注意：$n$ 个人的圆排列恰有 $n$ 个空隙，不是直线排列的 $n+1$ 个。）
把 A、B、C 安置在 $3$ 个不同空隙并排序。$\binom{7}{3}\cdot 3! = 35\cdot 6 = 210$ 种。
相乘。$720 \cdot 210 = 151\,200$。

圆上空隙 vs. 直线空隙。本题最常见的失误是套用"直线"的空隙数（那样会得 $7$ 人产生 $8$ 个空隙）。在圆上，$n$ 个元素恰产生 $n$ 个空隙，因为直线"接首尾" —— "最后一人之后"与"第一人之前"是同一个空隙。计数前，先在草稿上画一个 $4$–$5$ 个点的小圆自我验证。

Q10HARDPaper 21.11 Extended Binomial — Scaling Trick[7 marks]

Approximate $\sqrt{2}$ using the extended binomial expansion of $\sqrt{1+x}$, applying a scaling trick because $x = 1$ is outside the radius of convergence.使用 $\sqrt{1+x}$ 的广义二项展开逼近 $\sqrt{2}$；由于 $x = 1$ 在收敛半径外，需用换元技巧。

Answers:答案： (a) $\sqrt{1+x} \approx 1 + \tfrac{x}{2} - \tfrac{x^{2}}{8} + \tfrac{x^{3}}{16}$, valid for $|x| < 1$在 $|x| < 1$ 时成立 · (d) $\sqrt{2} \approx 1.41421$ (error $\lesssim 10^{-7}$)（误差 $\lesssim 10^{-7}$）

(a) Expansion of $\sqrt{1+x}$ M1·A1·A1

$\sqrt{1+x} = (1+x)^{1/2}$. The extended binomial theorem with $n = \tfrac{1}{2}$: $$ (1+x)^{1/2} \;=\; 1 + \tfrac{1}{2}x + \frac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2!}x^{2} + \frac{(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})}{3!}x^{3} + \cdots $$ Simplify each generalised coefficient:

$r = 2$: $\dfrac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2} = \dfrac{-\tfrac{1}{4}}{2} = -\dfrac{1}{8}.$
$r = 3$: $\dfrac{(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})}{6} = \dfrac{\tfrac{3}{8}}{6} = \dfrac{1}{16}.$

Therefore $$ \sqrt{1+x} \;\approx\; 1 + \tfrac{x}{2} - \tfrac{x^{2}}{8} + \tfrac{x^{3}}{16}, \qquad |x| < 1. $$

(b) Why $x = 1$ doesn't work R1

The series converges only when $|x| < 1$ (strict inequality). At $x = 1$ we are on the boundary — the radius-of-convergence condition is violated — so the series cannot be used to evaluate $\sqrt{2} = \sqrt{1 + 1}$ directly.

(c) The scaling identity A1

$\sqrt{2} = \tfrac{7}{5}\,\sqrt{1+r}$. Square both sides: $$ 2 \;=\; \tfrac{49}{25}\,(1+r) \quad\Longrightarrow\quad 1+r = \tfrac{50}{49} \quad\Longrightarrow\quad r = \tfrac{1}{49}. \;\;\checkmark $$ Note $|r| = \tfrac{1}{49} \ll 1$, so the series converges (and converges quickly).

(d) Approximation to $5$ dp M1·A1

Apply the four-term series to $x = \tfrac{1}{49}$: $$ \sqrt{1 + \tfrac{1}{49}} \;\approx\; 1 + \tfrac{1}{2\cdot 49} - \tfrac{1}{8\cdot 49^{2}} + \tfrac{1}{16\cdot 49^{3}}. $$ Compute term-by-term (GDC):

$\tfrac{1}{98} \approx 0.0102040816$
$\tfrac{1}{8\cdot 2401} = \tfrac{1}{19\,208} \approx -0.0000520617$
$\tfrac{1}{16\cdot 117\,649} = \tfrac{1}{1\,882\,384} \approx +0.0000005313$

$\sqrt{1+\tfrac{1}{49}} \approx 1 + 0.0102040816 - 0.0000520617 + 0.0000005313 \approx 1.01015255$. Multiply by $\tfrac{7}{5} = 1.4$: $$ \sqrt{2} \;\approx\; 1.4 \cdot 1.01015255 \;\approx\; 1.41421357. $$ To $5$ dp: $\boxed{\,\sqrt{2} \approx 1.41421\,}$. GDC value: $\sqrt{2} = 1.41421356\ldots \to 1.41421$ to $5$ dp. They agree to all $5$ decimal places. The actual error is $\lvert 1.41421357 - 1.41421356\rvert \approx 10^{-8}$ — far below $5$-dp precision.

Why the trick works — and where it generalises. The series for $(1+x)^{n}$ converges fast when $|x|$ is small. Direct evaluation at the “target” $x$ may lie outside the radius of convergence (or converge so slowly the truncation error swamps the answer). The trick is to factor the target as $C\cdot\sqrt{1 + r}$ where $C$ is a rational rational-approximation of the answer and $r$ is the tiny correction. Here $\tfrac{7}{5} = 1.4$ is the simple rational close to $\sqrt{2}$; the correction $r = \tfrac{1}{49}$ is small enough that even four series terms give $5$-dp accuracy. The same idea (using $\tfrac{17}{12}$ instead of $\tfrac{7}{5}$, with $r = -\tfrac{1}{288}$) gives $\sqrt{2}$ to $8$ dp. This is the standard route Paper 2 / Paper 3 questions take when the “obvious” substitution is outside the radius of convergence.

(a) $\sqrt{1+x}$ 的展开 M1·A1·A1

$\sqrt{1+x} = (1+x)^{1/2}$。在广义二项定理中取 $n = \tfrac{1}{2}$： $$ (1+x)^{1/2} \;=\; 1 + \tfrac{1}{2}x + \frac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2!}x^{2} + \frac{(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})}{3!}x^{3} + \cdots $$ 化简每个广义系数：

$r = 2$：$\dfrac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2} = \dfrac{-\tfrac{1}{4}}{2} = -\dfrac{1}{8}$。
$r = 3$：$\dfrac{(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2})}{6} = \dfrac{\tfrac{3}{8}}{6} = \dfrac{1}{16}$。

因此 $$ \sqrt{1+x} \;\approx\; 1 + \tfrac{x}{2} - \tfrac{x^{2}}{8} + \tfrac{x^{3}}{16}, \qquad |x| < 1. $$

(b) 为何 $x = 1$ 不可行 R1

级数仅在 $|x| < 1$（严格不等式）时收敛。$x = 1$ 恰位于边界 —— 收敛半径条件被破坏 —— 故不能用该级数直接求 $\sqrt{2} = \sqrt{1 + 1}$。

(c) 换元恒等式 A1

$\sqrt{2} = \tfrac{7}{5}\,\sqrt{1+r}$。两边平方： $$ 2 \;=\; \tfrac{49}{25}\,(1+r) \quad\Longrightarrow\quad 1+r = \tfrac{50}{49} \quad\Longrightarrow\quad r = \tfrac{1}{49}. \;\;\checkmark $$ 注意 $|r| = \tfrac{1}{49} \ll 1$，因此级数收敛且收敛极快。

(d) $5$ 位小数的逼近 M1·A1

对 $x = \tfrac{1}{49}$ 应用四项级数： $$ \sqrt{1 + \tfrac{1}{49}} \;\approx\; 1 + \tfrac{1}{2\cdot 49} - \tfrac{1}{8\cdot 49^{2}} + \tfrac{1}{16\cdot 49^{3}}. $$ 用 GDC 逐项算：

$\tfrac{1}{98} \approx 0.0102040816$
$\tfrac{1}{8\cdot 2401} = \tfrac{1}{19\,208} \approx -0.0000520617$
$\tfrac{1}{16\cdot 117\,649} = \tfrac{1}{1\,882\,384} \approx +0.0000005313$

$\sqrt{1+\tfrac{1}{49}} \approx 1 + 0.0102040816 - 0.0000520617 + 0.0000005313 \approx 1.01015255$。再乘以 $\tfrac{7}{5} = 1.4$： $$ \sqrt{2} \;\approx\; 1.4 \cdot 1.01015255 \;\approx\; 1.41421357. $$ 保留 $5$ 位小数：$\boxed{\,\sqrt{2} \approx 1.41421\,}$。GDC 值：$\sqrt{2} = 1.41421356\ldots \to 1.41421$。两者在 $5$ 位小数上完全一致。实际误差 $\lvert 1.41421357 - 1.41421356\rvert \approx 10^{-8}$ —— 远小于 $5$ 位小数精度。

为何该技巧有效，又能推广到何处。$(1+x)^{n}$ 的级数在 $|x|$ 很小时收敛很快。在"目标" $x$ 处直接代入可能落在收敛半径外（或收敛极慢，使截断误差淹没答案）。技巧是把目标分解为 $C\cdot\sqrt{1 + r}$，其中 $C$ 是接近目标值的简单有理数，$r$ 是微小的修正。这里 $\tfrac{7}{5} = 1.4$ 是接近 $\sqrt{2}$ 的简单有理数；修正项 $r = \tfrac{1}{49}$ 足够小，仅四项级数即可达到 $5$ 位小数精度。同样的思路（把 $\tfrac{7}{5}$ 换成 $\tfrac{17}{12}$，$r = -\tfrac{1}{288}$）可将 $\sqrt{2}$ 精确到 $8$ 位小数。Paper 2 / Paper 3 在"直接代入"超出收敛半径时，标准做法就是这个。

Q12HARDPaper 21.10 Permutation-Order Probability[8 marks]

Random permutation of $\{1, 2, \ldots, 10\}$ (uniform over $10!$ permutations). Find: (a) P($1$ and $2$ adjacent), (b) P($1$ before $2$), (c) P($1, 2, 3$ in increasing positional order), (d) P($k$ specified digits in increasing order).$\{1, 2, \ldots, 10\}$ 的随机排列（在 $10!$ 个排列上均匀分布）。求：(a) P($1$ 与 $2$ 相邻)，(b) P($1$ 在 $2$ 前)，(c) P($1, 2, 3$ 按位置递增)，(d) P(指定 $k$ 个数字按位置递增)。

Answers:答案： (a) $\dfrac{1}{5}$ · (b) $\dfrac{1}{2}$ · (c) $\dfrac{1}{6}$ · (d) $\dfrac{1}{k!}$

(a) $1$ and $2$ adjacent — block method M1·A1

Glue $1$ and $2$ together as a single super-element — this can be done in $2$ internal orders ("$12$" or "$21$"). The block plus the remaining $8$ digits gives $9$ units to permute: $9!$ arrangements. So: $$ P(\text{1 and 2 adjacent}) \;=\; \frac{2\cdot 9!}{10!} \;=\; \frac{2}{10} \;=\; \frac{1}{5}. $$

(b) $1$ to the left of $2$ — symmetry R1·A1

Consider any permutation $\pi$ and the permutation $\pi'$ obtained by swapping the positions of $1$ and $2$ (leaving every other digit fixed). The map $\pi \mapsto \pi'$ is a bijection from $\{\pi : 1\text{ left of }2\}$ to $\{\pi : 2\text{ left of }1\}$. These two sets are disjoint, partition all $10!$ permutations, and have equal size. Hence $$ P(\text{1 left of 2}) \;=\; \tfrac{1}{2}. $$

(c) $1, 2, 3$ in increasing positional order M1·A1

Fix any permutation; look only at the three positions occupied by $1$, $2$, $3$. Conditional on the choice of those three positions, the three digits land in those positions in one of $3! = 6$ orderings, each equally likely (uniform over $10!$ permutations $\Rightarrow$ uniform conditional). Exactly one of the $6$ orderings places them in increasing order. So $$ P(1\text{ left of }2\text{ left of }3) \;=\; \frac{1}{3!} \;=\; \frac{1}{6}. $$

(d) $k$ specified digits in increasing order R1·A1

The same symmetry argument: the relative order of any $k$ specific digits is uniformly distributed over the $k!$ orderings of those digits, regardless of which positions they occupy. Exactly $1$ of $k!$ orderings is the increasing one, so the probability is $$ \frac{1}{k!}. $$

The point. When the question asks about relative order of a small subset within a long random permutation, the long permutation is a red herring — the relative order of any $k$ items is uniform over the $k!$ orderings of those items, independent of the rest. This collapses problems of size $n!$ to size $k!$. The same idea drives Selection Sort / Quicksort average-case analyses and the "$95\%$ of comparisons" intuition behind randomised algorithms — once you know it, you'll see it in every "what's the chance a list is sorted on its first $k$ entries" question.

(a) $1$ 与 $2$ 相邻 —— 捆绑法 M1·A1

把 $1$ 与 $2$ 粘成一个超级元素 —— 内部顺序可为"$12$"或"$21$"，共 $2$ 种。该块与其余 $8$ 个数字共构成 $9$ 个单位，进行 $9!$ 种排列。故记事件 $A$ = $1$ 与 $2$ 相邻： $$ P(A) \;=\; \frac{2\cdot 9!}{10!} \;=\; \frac{2}{10} \;=\; \frac{1}{5}. $$

(b) $1$ 在 $2$ 的左侧 —— 对称性 R1·A1

考虑任意排列 $\pi$，以及把 $1$ 与 $2$ 位置互换（其余不变）得到的排列 $\pi'$。记 $L$ = "$1$ 在 $2$ 之左" 的排列集合，$L'$ = "$2$ 在 $1$ 之左" 的排列集合。映射 $\pi \mapsto \pi'$ 给出 $L \to L'$ 的双射。两集合互不相交，合起来是全部 $10!$ 个排列，且大小相等。因此 $$ P(L) \;=\; \tfrac{1}{2}. $$

(c) $1, 2, 3$ 按位置递增出现 M1·A1

固定任意排列；只看 $1$、$2$、$3$ 所占的三个位置。给定这三个位置后，三个数字落在其中的 $3! = 6$ 种顺序各以同样概率出现（在 $10!$ 上均匀 $\Rightarrow$ 条件分布也均匀）。$6$ 种中恰有 $1$ 种是递增排列。故（事件：$1 \to 2 \to 3$ 按位置递增出现。） $$ P(1 < 2 < 3 \text{ in position}) \;=\; \frac{1}{3!} \;=\; \frac{1}{6}. $$

(d) 指定 $k$ 个数字按位置递增 R1·A1

同样的对称性论证：任意 $k$ 个指定数字的相对顺序在它们的 $k!$ 种排列上均匀分布，与所占位置无关。$k!$ 中恰有 $1$ 种是递增的，所以概率为 $$ \frac{1}{k!}. $$

核心观察。当题目问的是长随机排列中某个小子集的相对顺序，长排列其实是一个"红鲱鱼"（干扰项） —— 任意 $k$ 个元素的相对顺序，在该 $k$ 个元素的 $k!$ 种顺序上均匀分布，与其余无关。这把规模 $n!$ 的问题缩到规模 $k!$。同一思想也驱动选择排序 / 快速排序的平均情形分析，以及随机算法中"$95\%$ 比较次数"那类直觉 —— 一旦掌握，你会在所有"前 $k$ 项已排好序的概率是多少"的问题里看到它。

Paper 3 — Worked Solutions第三卷 —— 详细解析

Q13HARDPaper 3AHL 1.10 Vandermonde & Sum-of-Squares[16 marks]

Develop Vandermonde's identity from Pascal's identity; apply to $m = n = 3$, $r = 2$ then to the squared-binomial sum. Six parts (a)–(f).从帕斯卡恒等式推出范德蒙恒等式；先在 $m = n = 3$、$r = 2$ 验证，再用于二项式系数平方和。六小问 (a)–(f)。

Answers:答案： (a) $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ · (b) $\text{both sides} = 15$ · (d) $\sum_{k=0}^{n}\binom{n}{k}^{\!2} = \binom{2n}{n}$ · (e) $252 = 252$ · (f) $\binom{2n}{n-1}$, with $m = n$, $r = n-1$

(a) Pascal's identity + committee interpretation A1·R1

$$ \binom{n}{r} \;=\; \binom{n-1}{r-1} + \binom{n-1}{r}. $$ Committee reading: to choose an $r$-person committee from $n$ people, fix one person (Alice). Either Alice is on the committee — pick the remaining $r-1$ from the other $n-1$ people, giving $\binom{n-1}{r-1}$ ways; or Alice is off — pick all $r$ from the other $n-1$, giving $\binom{n-1}{r}$ ways. Disjoint and exhaustive, so the two count $\binom{n}{r}$ between them.

(b) Verify $m = n = 3$, $r = 2$ M1·A1·A1

LHS: $\binom{6}{2} = \dfrac{6 \cdot 5}{2} = 15$. RHS: $$ \sum_{k=0}^{2} \binom{3}{k}\binom{3}{2-k} \;=\; \binom{3}{0}\binom{3}{2} + \binom{3}{1}\binom{3}{1} + \binom{3}{2}\binom{3}{0} \;=\; 1\cdot 3 + 3\cdot 3 + 3\cdot 1 \;=\; 3 + 9 + 3 \;=\; 15. $$ LHS $=$ RHS $= 15$. $\square$

(c) Combinatorial proof M1·M1·A1·R1

Let $S$ be a set of size $m + n$, partitioned into disjoint blocks $A$ of size $m$ and $B$ of size $n$.

LHS counts all $r$-element subsets of $S$ directly: $\binom{m+n}{r}$.
RHS counts the same family, classified by $k := |T \cap A|$. Given $k$, choose the $A$-part of $T$ in $\binom{m}{k}$ ways and (independently) the $B$-part of $T$ in $\binom{n}{r-k}$ ways. Summing over all valid $k \in \{0, 1, \ldots, r\}$ partitions the $r$-subsets of $S$ by their $A$-overlap.

Both sides count the same objects $\Rightarrow$ they are equal. $\square$

(d) Specialise $m = n$, $r = n$ M1·M1·A1

Vandermonde with $m = n$, $r = n$: $$ \binom{2n}{n} \;=\; \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}. $$ Symmetry $\binom{n}{n-k} = \binom{n}{k}$ collapses the product: $$ \binom{2n}{n} \;=\; \sum_{k=0}^{n} \binom{n}{k}\binom{n}{k} \;=\; \sum_{k=0}^{n} \binom{n}{k}^{\!2}. \qquad\square $$

(e) GDC check at $n = 5$ A1·A1

Coefficients $\binom{5}{0}, \binom{5}{1}, \ldots, \binom{5}{5} = 1, 5, 10, 10, 5, 1$. Squared and summed: $$ \sum_{k=0}^{5}\binom{5}{k}^{\!2} \;=\; 1 + 25 + 100 + 100 + 25 + 1 \;=\; 252. $$ And $\binom{10}{5} = \dfrac{10!}{5!\,5!} = 252$. Both equal $252$ — identity confirmed.

(f) Closed form for $\sum_{k=0}^{n-1} \binom{n}{k}\binom{n}{k+1}$ M1·A1

Apply Vandermonde with $m = n$, $r = n - 1$: $$ \binom{2n}{n-1} \;=\; \sum_{j=0}^{n-1} \binom{n}{j}\binom{n}{n-1-j}. $$ Symmetry: $\binom{n}{n-1-j} = \binom{n}{n - (n-1-j)} = \binom{n}{j+1}$. So $$ \sum_{j=0}^{n-1} \binom{n}{j}\binom{n}{j+1} \;=\; \binom{2n}{n-1}. $$ Choice: $m = n$, $r = n - 1$; symmetry step uses $\binom{n}{n-1-j} = \binom{n}{j+1}$.

Why Vandermonde matters. Pascal's identity is the special case $m = 1$ — the universe is partitioned into one element (Alice) plus the remaining $n - 1$. Vandermonde generalises this to any two-block partition, and the same "partition + classify by intersection size" recipe drives the entire Hockey Stick identity, the Chu–Vandermonde identity for upper non-integer index, and most generating-function combinatorics. Paper 3 mark-by-mark rewards: (i) stating the identity precisely before using it, (ii) writing the case-split clearly, (iii) doing one numeric sanity check before claiming a general result.

(a) 帕斯卡恒等式 + 委员会解释 A1·R1

$$ \binom{n}{r} \;=\; \binom{n-1}{r-1} + \binom{n-1}{r}. $$ 委员会解读：从 $n$ 人中选 $r$ 人组成委员会，先固定某一人（Alice）。要么 Alice 在委员会内 —— 再从其余 $n-1$ 人中选 $r-1$ 人，共 $\binom{n-1}{r-1}$ 种；要么 Alice 不在 —— 从其余 $n-1$ 人中选全部 $r$ 人，共 $\binom{n-1}{r}$ 种。两类不相交且穷尽全部情形，合计即 $\binom{n}{r}$。

(b) 验证 $m = n = 3$、$r = 2$ M1·A1·A1

左边：$\binom{6}{2} = \dfrac{6 \cdot 5}{2} = 15$。 右边： $$ \sum_{k=0}^{2} \binom{3}{k}\binom{3}{2-k} \;=\; \binom{3}{0}\binom{3}{2} + \binom{3}{1}\binom{3}{1} + \binom{3}{2}\binom{3}{0} \;=\; 1\cdot 3 + 3\cdot 3 + 3\cdot 1 \;=\; 15. $$ 左边 $=$ 右边 $= 15$。$\square$

(c) 组合证明 M1·M1·A1·R1

设集合 $S$ 含 $m + n$ 个元素，划分为大小为 $m$ 的块 $A$ 与大小为 $n$ 的块 $B$，二者不相交。

左边直接计数 $S$ 的所有 $r$ 元子集：$\binom{m+n}{r}$。
右边按 $k := |T \cap A|$ 分类同一族子集。给定 $k$，从 $A$ 中选 $T$ 的 $A$ 部分有 $\binom{m}{k}$ 种，独立地从 $B$ 中选 $r-k$ 个有 $\binom{n}{r-k}$ 种。对所有合法 $k \in \{0, 1, \ldots, r\}$ 求和，恰好按 $A$ 部分的大小将所有 $r$ 元子集分类完毕。

两边数的是同一族对象 $\Rightarrow$ 二者相等。$\square$

(d) 特殊化 $m = n$、$r = n$ M1·M1·A1

在范德蒙恒等式中代入 $m = n$、$r = n$： $$ \binom{2n}{n} \;=\; \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}. $$ 由对称性 $\binom{n}{n-k} = \binom{n}{k}$，乘积合并为平方： $$ \binom{2n}{n} \;=\; \sum_{k=0}^{n} \binom{n}{k}\binom{n}{k} \;=\; \sum_{k=0}^{n} \binom{n}{k}^{\!2}. \qquad\square $$

(e) $n = 5$ 的 GDC 数值核对 A1·A1

系数 $\binom{5}{0}, \binom{5}{1}, \ldots, \binom{5}{5} = 1, 5, 10, 10, 5, 1$。平方求和： $$ \sum_{k=0}^{5}\binom{5}{k}^{\!2} \;=\; 1 + 25 + 100 + 100 + 25 + 1 \;=\; 252. $$ 又 $\binom{10}{5} = \dfrac{10!}{5!\,5!} = 252$。两值同为 $252$，恒等式得到验证。

(f) $\sum_{k=0}^{n-1} \binom{n}{k}\binom{n}{k+1}$ 的闭合形式 M1·A1

在范德蒙恒等式中取 $m = n$、$r = n - 1$： $$ \binom{2n}{n-1} \;=\; \sum_{j=0}^{n-1} \binom{n}{j}\binom{n}{n-1-j}. $$ 对称性：$\binom{n}{n-1-j} = \binom{n}{n - (n-1-j)} = \binom{n}{j+1}$。故 $$ \sum_{j=0}^{n-1} \binom{n}{j}\binom{n}{j+1} \;=\; \binom{2n}{n-1}. $$ 选取：$m = n$、$r = n - 1$；所用对称性步骤为 $\binom{n}{n-1-j} = \binom{n}{j+1}$。

范德蒙的意义。帕斯卡恒等式是 $m = 1$ 的特例 —— 把全集划分为"一个元素（Alice）"与"剩余 $n - 1$ 个"。范德蒙将其推广到任意双块划分，同样的"分块 + 按交集大小分类"双计数法则贯穿曲棍球棒恒等式（Hockey Stick）、上指标非整数的 Chu–Vandermonde 推广，乃至整个生成函数组合学。Paper 3 逐分给分的重点：(i) 在使用之前精确写出恒等式；(ii) 把分类讨论写清楚；(iii) 在断言一般结论前做一次数值核对。

Combinatorics — Solutions组合数学 —— 解析