Fix the mistake: Counting Cases ① (Permutations)
Counting① Sets and the Basics of Counting
Counting problems concern the skill of accurately determining “how many possibilities there are.” Learn the principles for counting without duplication or omission.
Addition Rule and Multiplication Rule
- Addition rule: Number of cases in which A or B occurs → m + n cases (when mutually exclusive)
- Multiplication rule: Number of cases in which A and B occur consecutively → m × n cases
Example ①
If there are 3 ways to travel from Point A to Point B and 4 ways to travel from Point B to Point C, how many ways are there to travel A→B→C?
Solution: By the multiplication rule, 3 × 4 = 12 ways
If there are 3 ways to travel from Point A to Point B and 4 ways to travel from Point B to Point C, how many ways are there to travel A→B→C?
Solution: By the multiplication rule, 3 × 4 = 12 ways
Inclusion–Exclusion Principle
|A ∪ B| = |A| + |B| − |A ∩ B|
Example ②
Among the integers from 1〜30, find the number of multiples of 3 or 5.
Solution: 10 multiples of 3 + 6 multiples of 5 − 2 multiples of 15 = 14
Among the integers from 1〜30, find the number of multiples of 3 or 5.
Solution: 10 multiples of 3 + 6 multiples of 5 − 2 multiples of 15 = 14
Key Point
- “Or” → addition rule (watch for overlap)
- “And (consecutively)” → multiplication rule
- Overlapping “or” → inclusion–exclusion principle
Practice Problems
- Flip a coin and roll a die simultaneously. How many outcomes are there?
- Among the integers from 1〜20, find the number of multiples of 2 or 3.
- There are 5 ways from Point A to Point B, 3 ways from Point B to Point C, and 2 direct ways from Point A to Point C. How many ways are there from A→C in total?
Solution and Explanation
- Answer: 2 × 6 = 12 ways
- Answer: 10 multiples of 2 + 6 multiples of 3 − 3 multiples of 6 = 13
- Answer: 5×3 + 2 = 17 ways
Previous and next lessons in this course
- Previous lesson: Fix the mistake: Counting Cases ① (Permutations)
- Next lesson: Fix the mistake: Counting Cases ① (Permutations)