🔢 Counting Principles

The three fundamental rules that form the foundation of all counting problems.

🎯 Key Ideas

Addition Principle

When counting objects that can be categorized into mutually exclusive groups, add the counts of each group.

Formula: If $A$ and $B$ are disjoint sets, then $|A \cup B| = |A| + |B|$

Multiplication Principle

When counting sequences of choices where each choice is independent, multiply the number of choices at each step.

Formula: If you can make $m$ choices for the first decision and $n$ choices for the second decision, then there are $m \cdot n$ total ways to make both decisions.

Complement Principle

Instead of counting what you want, count what you don’t want and subtract from the total.

Formula: $|A| = |S| - |A^c|$ where $S$ is the universal set.

đź’ˇ Micro-Examples

Addition Principle

  • Problem: How many ways can you choose a red or blue ball from a bag with 3 red, 4 blue, and 2 green balls?
  • Solution: $3 + 4 = 7$ ways (red OR blue)

Multiplication Principle

  • Problem: How many ways can you choose a shirt and pants if you have 3 shirts and 4 pants?
  • Solution: $3 \cdot 4 = 12$ ways (shirt AND pants)

Complement Principle

  • Problem: How many ways can you roll two dice and get at least one 6?
  • Solution: Total ways = $6^2 = 36$, ways with no 6s = $5^2 = 25$, so answer = $36 - 25 = 11$

⚠️ Common Traps & Fixes

Trap: Using addition when you should use multiplication

  • Wrong: “I can take the bus (3 routes) or walk (2 routes), so there are $3 + 2 = 5$ ways to get to school”
  • Right: If you’re counting the total number of ways to get to school, it’s $3 + 2 = 5$. But if you’re counting ways to get to school AND back home, it’s $(3 + 2) \cdot (3 + 2) = 25$ ways.

Trap: Forgetting that sets must be disjoint for addition

  • Wrong: “There are 10 math students and 8 science students, so there are $10 + 8 = 18$ students total”
  • Right: If some students take both subjects, use inclusion-exclusion: $10 + 8 - |\text{math} \cap \text{science}|$

Trap: Overcounting with complement

  • Wrong: “At least 2 heads in 3 flips” = Total - “No heads” - “1 head” = $8 - 1 - 3 = 4$
  • Right: “At least 2 heads” = “Exactly 2 heads” + “Exactly 3 heads” = $3 + 1 = 4$

🏆 AMC-Style Worked Example

Problem: How many 3-digit numbers contain at least one 7?

Solution:

  • Total 3-digit numbers: $9 \cdot 10 \cdot 10 = 900$ (first digit: 1-9, others: 0-9)
  • 3-digit numbers with NO 7s: $8 \cdot 9 \cdot 9 = 648$ (first digit: 1-6,8,9; others: 0-6,8,9)
  • Answer: $900 - 648 = 252$ numbers

Key insight: Use complement counting to avoid casework on “exactly 1 seven”, “exactly 2 sevens”, etc.

đź”— Related Topics

Next: Permutations & Combinations → Inclusion-Exclusion