14.1 Fundamental Counting Principles (Addition and Multiplication Rules)

14.1 Fundamental Counting Principles (Addition and Multiplication Rules)

Counting principles provide the foundational mathematical framework for discrete mathematics, combinatorics, and probability theory in Gaokao Mathematics. Every complex counting problem—no matter how elaborate—can be reduced to systematic applications of two core rules: the Addition Principle (Rule of Sum) and the Multiplication Principle (Rule of Product). Mastering the distinction between when to add and when to multiply is essential for solving high-level Gaokao problems without overcounting (重复) or undercounting (遗漏).

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1. Mathematical Definitions and Formal Foundations

1.1 The Addition Principle (Rule of Sum / 分类加法计数原理)

If a task $T$ can be accomplished by choosing from one of $k$ mutually exclusive and exhaustive categories (or cases) $C_1, C_2, \dots, C_k$, where category $C_1$ contains $m_1$ distinct ways to complete the task, category $C_2$ contains $m_2$ distinct ways, ..., and category $C_k$ contains $m_k$ distinct ways, then the total number of distinct ways to complete task $T$ is given by:

$N = m_1 + m_2 + \dots + m_k = \sum_{i=1}^k m_i$

Set-Theoretic Interpretation: Let $S_1, S_2, \dots, S_k$ be pairwise disjoint sets representing the sets of options for each category, meaning $S_i \cap S_j = \emptyset$ for all $i \neq j$. The total set of choices is the disjoint union $S = \bigcup_{i=1}^k S_i$, and its cardinality is:

$|S| = |S_1| + |S_2| + \dots + |S_k|$

Key Criterion for Addition: The choices in different categories are mutually exclusive (互斥). Selecting a choice from category $C_i$ completes the task independently of the choices in any other category $C_j$.

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1.2 The Multiplication Principle (Rule of Product / 分步乘法计数原理)

If a task $T$ must be accomplished through a sequence of $k$ consecutive steps $S_1, S_2, \dots, S_k$, where Step 1 can be completed in $m_1$ distinct ways, Step 2 can be completed in $m_2$ distinct ways (regardless of the outcome of Step 1), ..., and Step $k$ can be completed in $m_k$ distinct ways (regardless of the outcomes of prior steps), then the total number of distinct ways to complete task $T$ is given by:

$N = m_1 \times m_2 \times \dots \times m_k = \prod_{i=1}^k m_i$

Set-Theoretic Interpretation: If $A_1, A_2, \dots, A_k$ are the sets of options available at each stage, the total set of multi-step outcomes is represented by the Cartesian product $A = A_1 \times A_2 \times \dots \times A_k$. The cardinality is:

$|A| = |A_1| \times |A_2| \times \dots \times |A_k|$

Key Criterion for Multiplication: The task is incomplete after any single step; all steps must be executed sequentially in order to achieve a complete outcome.

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1.3 Strategic Comparison: Classification vs. Step-by-Step Decomposition

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2. Advanced Problem-Solving Methodologies for Gaokao

In Gaokao examinations, counting problems rarely present straightforward single-step calculations. Instead, students must structure complex conditions into rigorous mathematical schemes using three primary analytical methodologies:

2.1 The Exhaustive Case-Classification Strategy (分类讨论法)

When a problem contains multiple coupled constraints (such as restricted positions, specific digit properties, or color adjacency rules), classify the sample space into sub-cases based on a primary restricting variable.

1. Rule of Non-Overlapping & Non-Omission (不重不漏):
   • The union of all sub-cases must equal the entire valid space: $\bigcup_{i=1}^k C_i = \mathcal{S}_{\text{valid}}$.
   • No two sub-cases may share an outcome: $C_i \cap C_j = \emptyset$ for $i \neq j$.
2. Choosing the Classification Standard (选定分类标准): Always select the most restrictive condition as the branching variable (e.g., whether a specific region uses the same color as another non-adjacent region, or whether a unit digit is zero vs. non-zero).

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2.2 The Indirect / Complementary Subtractive Strategy (正难则反 - 间接法)

When direct calculation requires analyzing numerous tedious cases (such as conditions involving "at least one", "not all equal", or "at least two adjacent"), switch to complementary counting:

$N(\text{Satisfying Conditions}) = N(\text{Total Sample Space}) - N(\text{Violating Conditions})$

Example Principle: To count the number of 4-digit numbers containing at least one digit '8', it is far simpler to subtract the number of 4-digit numbers containing no '8' from the total number of valid 4-digit numbers:

$N = (9 \times 10^3) - (8 \times 9^3) = 9000 - 5832 = 3168$

3. Classic Gaokao Worked Examples and Solutions

Worked Example 1: Map and Region Coloring under Adjacency Constraints

Problem Statement: A map consists of 5 regions labeled $A, B, C, D, E$. Region $A$ is adjacent to all four other regions ($B, C, D, E$). Region $B$ is adjacent to $A, C, E$. Region $C$ is adjacent to $A, B, D$. Region $D$ is adjacent to $A, C, E$. Region $E$ is adjacent to $A, B, D$. There are $4$ distinct colors available. Adjacent regions must be painted in different colors. Calculate the total number of valid coloring schemes.

Detailed Mathematical Solution:

• Step 1: Analyze Adjacencies and Constraints. Region $A$ touches all other 4 regions ($B, C, D, E$). Therefore, region $A$ must be assigned a unique color first.

• Step 2: Assign Color to Region $A$. Since there are $4$ colors available, $A$ can be colored in $4$ ways. After coloring $A$, exactly $3$ colors remain for the rest of the regions $B, C, D, E$.

• Step 3: Classify based on non-adjacent regions $B$ and $D$. Regions $B$ and $D$ are not adjacent to each other. We divide into two mutually exclusive cases based on whether $B$ and $D$ receive the same color or different colors.

• Case 1: Regions $B$ and $D$ have the SAME color.

  1. Select color for $B$: There are $3$ available colors (excluding $A$'s color).
  2. Color for $D$: Since $D$ must share $B$'s color, there is only $1$ choice for $D$.
  3. Color for $C$: Region $C$ is adjacent to $A, B, D$. Since $B$ and $D$ share the same color, $C$ is adjacent to only $2$ distinct colors (the color of $A$ and the common color of $B, D$). Thus, $C$ has $4 - 2 = 2$ choices.
  4. Color for $E$: Region $E$ is adjacent to $A, B, D$. Similarly, $E$ has $4 - 2 = 2$ choices.
  • Number of ways for Case 1: $N_1 = 3 \times 1 \times 2 \times 2 = 12$ ways.

• Case 2: Regions $B$ and $D$ have DIFFERENT colors.

  1. Select color for $B$: $3$ choices (excluding $A$'s color).
  2. Select color for $D$: $D$ cannot use $A$'s color nor $B$'s color, leaving $4 - 2 = 2$ choices.
  3. Color for $C$: Region $C$ is adjacent to $A, B, D$. Since $A, B, D$ all have mutually distinct colors, $3$ colors are used up. $C$ has $4 - 3 = 1$ choice.
  4. Color for $E$: Region $E$ is adjacent to $A, B, D$. Since $A, B, D$ have distinct colors, $E$ has $4 - 3 = 1$ choice.
  • Number of ways for Case 2: $N_2 = 3 \times 2 \times 1 \times 1 = 6$ ways.

• Step 4: Combine Cases and Multiply by $A$'s choices. The sub-total choices for regions $(B, C, D, E)$ is $N_1 + N_2 = 12 + 6 = 18$. Multiplying by the choices for region $A$: $N_{\text{total}} = 4 \times (12 + 6) = 4 \times 18 = 72$

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Worked Example 2: Constrained Digit Composition

Problem Statement: Using the digits ${0, 1, 2, 3, 4, 5}$, how many 4-digit even numbers without repeated digits can be formed?

Detailed Mathematical Solution:

• Step 1: Identify Constraints.

  1. A 4-digit number cannot have $0$ in the thousands place (thousands digit $\neq 0$).
  2. An even number must have an even units digit ($0, 2,$ or $4$).
  3. Digits cannot be repeated (without replacement).

• Step 2: Classify based on the Units Digit. Since $0$ affects both the thousands place constraint and the even units place constraint, we split into two mutually exclusive cases depending on whether the units digit is $0$ or non-zero even ($2$ or $4$).

• Case 1: The units digit is $0$.

  • Units digit: $1$ choice (digit $0$).
  • Thousands digit: Can be any of ${1, 2, 3, 4, 5}$, so $5$ choices.
  • Hundreds digit: Select from the remaining $4$ digits, so $4$ choices.
  • Tens digit: Select from the remaining $3$ digits, so $3$ choices.
  • Number of ways for Case 1: $N_1 = 1 \times 5 \times 4 \times 3 = 60$

• Case 2: The units digit is $2$ or $4$.

  • Units digit: $2$ choices (either $2$ or $4$).
  • Thousands digit: Must not be $0$ and must not be the units digit chosen. Out of $6$ digits, excluding $0$ and the chosen units digit leaves $6 - 2 = 4$ choices.
  • Hundreds digit: Can now include $0$, excluding the thousands and units digits. Remaining choices: $6 - 2 = 4$ choices.
  • Tens digit: Remaining choices: $6 - 3 = 3$ choices.
  • Number of ways for Case 2: $N_2 = 2 \times 4 \times 4 \times 3 = 96$

• Step 3: Apply the Addition Principle. $N_{\text{total}} = N_1 + N_2 = 60 + 96 = 156$