14.2 Permutations and Combinations

14.2 Permutations and Combinations

Permutations and combinations form the operational core of combinatorial counting in Gaokao Mathematics. While fundamental counting principles dictate when to add or multiply, permutation and combination formulas provide efficient algebraic tools to count ordered arrangements and unordered selections.

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1. Mathematical Definitions and Formulas

1.1 Permutations (排列)

A permutation of $n$ distinct elements taken $m$ at a time ($0 \le m \le n$) is an ordered arrangement of $m$ elements chosen from the set of $n$ elements. The total number of permutations is denoted by $A_n^m$ (or $P(n, m)$ or $P_n^m$) and is calculated by:

$A_n^m = n(n-1)(n-2)\cdots(n-m+1) = \frac{n!}{(n-m)!}$

Special Cases:

• When $m = n$, it represents a full permutation (全排列): $A_n^n = n! = n \times (n-1) \times \dots \times 2 \times 1$.
• By definition, $0! = 1$, which ensures $A_n^0 = \frac{n!}{n!} = 1$.

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1.2 Combinations (组合)

A combination of $n$ distinct elements taken $m$ at a time ($0 \le m \le n$) is an unordered selection of $m$ elements chosen from the set of $n$ elements. The total number of combinations is denoted by $C_n^m$ (or $\binom{n}{m}$) and is calculated by:

$C_n^m = \frac{A_n^m}{m!} = \frac{n!}{m!(n-m)!}$

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1.3 Key Combinatorial Identities and Derivations

1. Symmetry Property (对称性): $C_n^m = C_n^{n-m}$ Combinatorial Proof: Choosing $m$ elements to include from $n$ available items is logically equivalent to choosing $n-m$ elements to leave behind.

2. Pascal's Identity / Recurrence Relation (加法公式 / 杨辉等式): $C_n^m + C_n^{m-1} = C_{n+1}^m \quad (1 \le m \le n)$ Combinatorial Proof: Consider a set of $n+1$ elements including a distinguished element $x$. To choose $m$ elements from $n+1$ elements, split into two mutually exclusive cases:

   • Case 1: Element $x$ is included. We must choose the remaining $m-1$ elements from the other $n$ elements $\implies C_n^{m-1}$ ways.
   • Case 2: Element $x$ is excluded. We must choose all $m$ elements from the remaining $n$ elements $\implies C_n^m$ ways. Adding the two cases gives $C_n^m + C_n^{m-1} = C_{n+1}^m$.

3. Absorption / Reduction Identity: $k C_n^k = n C_{n-1}^{k-1} \implies C_n^k = \frac{n}{k} C_{n-1}^{k-1}$ Combinatorial Proof: Selecting a committee of $k$ people from $n$ people and designating 1 leader is equivalent to picking the leader first ($n$ choices) and then picking the remaining $k-1$ members from $n-1$ people ($C_{n-1}^{k-1}$ choices).

4. Sum of Combinations (Total Subsets): $\sum_{k=0}^n C_n^k = C_n^0 + C_n^1 + C_n^2 + \dots + C_n^n = 2^n$

2. Standard Problem-Solving Strategies for Constrained Counting

Gaokao permutation and combination problems frequently introduce spatial, positional, or structural constraints. Five canonical methods govern these problems:

2.1 The Bundle Method for Adjacent Elements (捆绑法)

Applicability: Used when specified elements must be adjacent (必须相邻). Procedure:

1. Treat the constrained adjacent elements as a single "bundled super-element".
2. Arrange the super-element along with all remaining unconstrained elements.
3. Multiply by the internal permutations of the elements within the bundle.

Formula Structure: If $k$ elements out of $n$ must be adjacent, total arrangements equal: $N = A_{n-k+1}^{n-k+1} \times A_k^k = (n-k+1)! \times k!$

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2.2 The Gap Insertion Method for Non-Adjacent Elements (插空法)

Applicability: Used when specified elements cannot be adjacent (不能相邻). Procedure:

1. Arrange all unconstrained elements first, creating $m+1$ available "gaps" (including the two ends) around $m$ arranged items.
2. Insert the $k$ constrained non-adjacent elements into the available gaps ($k \le m+1$).

Formula Structure: $N = A_m^m \times A_{m+1}^k = m! \times \frac{(m+1)!}{(m+1-k)!}$

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2.3 Priority Assignment for Restricted Positions/Elements (优先法)

Applicability: Used when specific elements must occupy (or avoid) specific positions. Procedure: First satisfy the strict spatial/elemental constraints before arranging or selecting the remaining unrestricted items.

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2.4 Grouping and Distribution Strategies (分组与分配问题)

Dividing $n$ distinct objects into $k$ groups is a core Gaokao topic:

1. Unequal Unnamed Grouping (不均匀分组): Dividing $n$ objects into groups of sizes $m_1, m_2, \dots, m_k$ (where all $m_i$ are distinct and $\sum m_i = n$): $N = C_n^{m_1} \times C_{n-m_1}^{m_2} \times \dots \times C_{m_k}^{m_k}$

2. Equal Unnamed Grouping (均匀分组): Dividing $n$ objects into $p$ groups of equal size $m$ ($n = p \times m$). Since the groups are unnamed and identical in size, multiplying combinations overcounts by the $p!$ permutations of the groups themselves: $N = \frac{C_n^m \times C_{n-m}^m \times \dots \times C_m^m}{p!}$

3. Distribution to Named Targets (定向分配): If the groups are subsequently assigned to $p$ distinct entities (e.g., $p$ different schools or cities), multiply the grouping count by $p!$: $N_{\text{named distribution}} = N_{\text{grouping}} \times p!$

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2.5 The Stars and Bars Method (隔板法)

Applicability: Finding the number of positive integer solutions to $x_1 + x_2 + \dots + x_k = n$ where $x_i \ge 1$. Procedure: Place $n$ identical items in a line, creating $n-1$ spaces between them. Insert $k-1$ dividers (bars) into these $n-1$ spaces. Formula: $N = C_{n-1}^{k-1}$ If $x_i \ge 0$ (non-negative integer solutions), substitute $y_i = x_i + 1 \implies y_1 + y_2 + \dots + y_k = n + k$ with $y_i \ge 1$: $N = C_{n+k-1}^{k-1}$

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3. Classic Gaokao Worked Examples and Solutions

Worked Example 1: Combined Adjacent and Non-Adjacent Restrictions

Problem Statement: Seven students $A, B, C, D, E, F, G$ stand in a single line for a photo. Calculate the number of valid line-up arrangements under the following simultaneous conditions:

1. Students $A$ and $B$ must stand adjacent to each other.
2. Students $C$ and $D$ cannot stand adjacent to each other.

Detailed Mathematical Solution:

• Step 1: Bundle the adjacent students $A$ and $B$. Treat $(A, B)$ as a single bundled block $X$. Inside block $X$, $A$ and $B$ can be arranged in $A_2^2 = 2! = 2$ ways.
• Step 2: Identify the remaining unconstrained entities. The entities to be arranged first (excluding $C$ and $D$) are block $X$, student $E$, student $F$, and student $G$. This gives $4$ items in total.
• Step 3: Arrange the unconstrained entities to form gaps for $C$ and $D$. Arranging these $4$ items ($X, E, F, G$) in a row can be done in: $A_4^4 = 4! = 24 \text{ ways}$
• Step 4: Count available gaps for $C$ and $D$. Arranging $4$ items creates $4 + 1 = 5$ available gaps (including both ends): $\underline{\quad} \, \text{Item}_1 \, \underline{\quad} \, \text{Item}_2 \, \underline{\quad} \, \text{Item}_3 \, \underline{\quad} \, \text{Item}_4 \, \underline{\quad}$
• Step 5: Insert non-adjacent students $C$ and $D$ into the gaps. We choose $2$ distinct gaps out of $5$ for $C$ and $D$ in order: $A_5^2 = 5 \times 4 = 20 \text{ ways}$
• Step 6: Multiply all stage counts using the Multiplication Principle. $N_{\text{total}} = A_2^2 \times A_4^4 \times A_5^2 = 2 \times 24 \times 20 = 960$

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Worked Example 2: Grouping and Assigned Distribution

Problem Statement: Six distinct volunteer teachers are to be assigned to $3$ different rural schools ($S_1, S_2, S_3$) such that each school receives at least $1$ teacher. How many different distribution schemes are possible?

Detailed Mathematical Solution:

• Step 1: Analyze integer partitions of $6$ into $3$ positive parts. The number of teachers $6$ can be partitioned into $3$ positive integer parts in three distinct ways:

  • Pattern 1: Group sizes $(4, 1, 1)$
  • Pattern 2: Group sizes $(3, 2, 1)$
  • Pattern 3: Group sizes $(2, 2, 2)$

• Case 1: Grouping pattern $(4, 1, 1)$.

  • Unnamed Grouping: Since two groups have equal size ($1, 1$), divide by $2!$: $N_{\text{group}, 1} = \frac{C_6^4 \times C_2^1 \times C_1^1}{2!} = \frac{15 \times 2 \times 1}{2} = 15$
  • Distribution to $3$ distinct schools: Multiply by $3!$: $N_{\text{dist}, 1} = 15 \times 3! = 15 \times 6 = 90$

• Case 2: Grouping pattern $(3, 2, 1)$.

  • Unnamed Grouping: All group sizes are distinct ($3, 2, 1$), no division required: $N_{\text{group}, 2} = C_6^3 \times C_3^2 \times C_1^1 = 20 \times 3 \times 1 = 60$
  • Distribution to $3$ distinct schools: Multiply by $3!$: $N_{\text{dist}, 2} = 60 \times 3! = 60 \times 6 = 360$

• Case 3: Grouping pattern $(2, 2, 2)$.

  • Unnamed Grouping: All three groups have equal size ($2, 2, 2$), divide by $3!$: $N_{\text{group}, 3} = \frac{C_6^2 \times C_4^2 \times C_2^2}{3!} = \frac{15 \times 6 \times 1}{6} = 15$
  • Distribution to $3$ distinct schools: Multiply by $3!$: $N_{\text{dist}, 3} = 15 \times 3! = 15 \times 6 = 90$

• Step 2: Sum all cases using the Addition Principle. $N_{\text{total}} = N_{\text{dist}, 1} + N_{\text{dist}, 2} + N_{\text{dist}, 3} = 90 + 360 + 90 = 540$