Lesson 2: Counting Methods
Module 3: Probability Basics • Lesson 2 of 4 • ~35-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Apply the Fundamental Counting Principle to count outcomes
- Calculate factorials and understand their use in counting
- Use permutations when order matters
- Use combinations when order doesn't matter
- Decide which counting method to use for different situations
Why Counting Methods Matter
In Lesson 1, we used P(A) = n(A) / n(S) to calculate probabilities. But what if the sample space has hundreds or thousands of outcomes? Listing them all is impractical!
Counting methods let us systematically count outcomes without listing every single one. This is essential for:
- Calculating probabilities in complex situations
- Analyzing games of chance (poker hands, lottery odds)
- Understanding statistical sampling
- Solving real-world optimization problems
Fundamental Counting Principle
Fundamental Counting Principle (FCP)
If one event can occur in m ways and a second event can occur in n ways, then the two events together can occur in m × n ways.
Extended: If there are k events that can occur in n₁, n₂, ..., n_k ways respectively, then the total number of ways all k events can occur is:
n₁ × n₂ × n₃ × ... × n_k
Example 1: Outfit Combinations
Question: You have 4 shirts and 3 pairs of pants. How many different outfits can you make?
Solution:
- Choose a shirt: 4 ways
- Choose pants: 3 ways
- Total outfits = 4 × 3 = 12
Example 2: License Plates
Question: A license plate consists of 2 letters followed by 3 digits. How many different plates are possible?
Solution:
- First letter: 26 choices (A-Z)
- Second letter: 26 choices
- First digit: 10 choices (0-9)
- Second digit: 10 choices
- Third digit: 10 choices
- Total = 26 × 26 × 10 × 10 × 10 = 676,000 plates
Example 3: Restaurant Menu
Question: A restaurant offers 5 appetizers, 8 main courses, and 4 desserts. How many different three-course meals can you order?
Solution:
Total meals = 5 × 8 × 4 = 160 possible three-course meals
Factorials
Factorial (n!)
The factorial of a positive integer n, written as n!, is the product of all positive integers from 1 to n.
n! = n × (n−1) × (n−2) × ... × 3 × 2 × 1
Special case: 0! = 1 (by definition)
Examples of Factorials:
- 3! = 3 × 2 × 1 = 6
- 4! = 4 × 3 × 2 × 1 = 24
- 5! = 5 × 4 × 3 × 2 × 1 = 120
- 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800
Note: Factorials grow very rapidly! Use a calculator for factorials larger than 5!
Calculator Tip
Most calculators have an "n!" or "x!" button. Use it for large factorials!
Example: To calculate 10!, press: 10, then the factorial button
Permutations (Order Matters)
Permutation
A permutation is an arrangement of objects where order matters.
Example: Arranging 3 books on a shelf. The order "ABC" is different from "BAC".
Permutations of n Objects
Number of ways to arrange n distinct objects:
n!
Example 4: Arranging Books
Question: How many ways can you arrange 5 different books on a shelf?
Solution:
Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
Permutations of r Objects from n (P(n,r) or nPr)
Number of ways to arrange r objects chosen from n distinct objects:
P(n,r) = n! / (n−r)!
Alternative notation: nPr or nPr
Example 5: Race Podium
Question: In a race with 10 runners, how many ways can gold, silver, and bronze medals be awarded?
Solution:
We're choosing 3 runners from 10, and order matters (gold ≠ silver ≠ bronze)
P(10,3) = 10! / (10−3)! = 10! / 7!
= (10 × 9 × 8 × 7!) / 7!
= 10 × 9 × 8 = 720 ways
Example 6: Password Creation
Question: How many 4-letter passwords can be formed from the letters A through F if no letter can be repeated?
Solution:
We're arranging 4 letters from 6, and order matters
P(6,4) = 6! / (6−4)! = 6! / 2!
= (6 × 5 × 4 × 3 × 2!) / 2!
= 6 × 5 × 4 × 3 = 360 passwords
Shortcut for Permutations
Instead of calculating the full factorial, just multiply the first few terms:
P(n,r) = n × (n−1) × (n−2) × ... × (n−r+1)
Example: P(10,3) = 10 × 9 × 8 = 720
Combinations (Order Doesn't Matter)
Combination
A combination is a selection of objects where order does NOT matter.
Example: Choosing 3 people for a committee. The group {Alice, Bob, Carol} is the same as {Carol, Alice, Bob}.
Number of ways to choose r objects from n distinct objects (order doesn't matter):
C(n,r) = n! / (r! × (n−r)!)
Alternative notation: nCr, nCr, or "n choose r"
Relationship Between Permutations and Combinations
C(n,r) = P(n,r) / r!
Why? When order doesn't matter, we divide by r! to remove duplicate arrangements.
Example 7: Committee Selection
Question: How many ways can you choose 3 people from a group of 8 to form a committee?
Solution:
Order doesn't matter (committee has no ranks), so we use combinations
C(8,3) = 8! / (3! × 5!)
= (8 × 7 × 6 × 5!) / (3! × 5!)
= (8 × 7 × 6) / (3 × 2 × 1)
= 336 / 6 = 56 ways
Example 8: Lottery
Question: In a lottery, you must choose 6 numbers from 1 to 49. How many different combinations are possible?
Solution:
Order doesn't matter, so we use combinations
C(49,6) = 49! / (6! × 43!)
= (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1)
= 10,068,347,520 / 720 = 13,983,816 combinations
(That's why winning the lottery is so unlikely!)
Example 9: Choosing Toppings
Question: A pizza shop offers 10 toppings. How many ways can you choose exactly 4 toppings?
Solution:
Order doesn't matter (same pizza regardless of topping order)
C(10,4) = 10! / (4! × 6!)
= (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
= 5,040 / 24 = 210 ways
Permutations vs. Combinations: Decision Guide
Which Method Should I Use?
| Question | Does Order Matter? | Method | Formula |
|---|---|---|---|
| Race winners (1st, 2nd, 3rd) | YES | Permutation | P(n,r) |
| Committee members | NO | Combination | C(n,r) |
| Password letters | YES | Permutation | P(n,r) |
| Lottery numbers | NO | Combination | C(n,r) |
| Seating arrangement | YES | Permutation | n! or P(n,r) |
| Selecting items from a menu | NO | Combination | C(n,r) |
Key Question to Ask:
"If I swap two items, do I get a different outcome?"
- YES (different outcome) → Order matters → Permutation
- NO (same outcome) → Order doesn't matter → Combination
Check Your Understanding
Practice Question 1
A restaurant has 3 appetizers, 5 entrees, and 4 desserts. How many different complete meals can be ordered?
Solution:
Use Fundamental Counting Principle:
Total meals = 3 × 5 × 4 = 60 complete meals
Practice Question 2
Calculate 6!
Solution:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Practice Question 3
How many ways can 4 students be arranged in a row for a photo?
Solution:
Order matters (different positions = different photo)
Number of arrangements = 4! = 24 ways
Practice Question 4
From a class of 12 students, how many ways can a teacher choose a president, vice president, and secretary?
Solution:
Order matters (president ≠ vice president ≠ secretary)
Use permutation: P(12,3) = 12 × 11 × 10 = 1,320 ways
Practice Question 5
How many ways can you choose 5 cards from a standard 52-card deck?
Solution:
Order doesn't matter (a hand is the same regardless of order dealt)
Use combination: C(52,5) = 52! / (5! × 47!)
= (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1)
= 311,875,200 / 120 = 2,598,960 ways
Practice Question 6
Should you use a permutation or combination? You're selecting 3 books from 10 to take on vacation.
Answer: Combination
Reasoning:
Order doesn't matter. Whether you choose Book A, Book B, Book C or Book C, Book A, Book B, you're taking the same 3 books.
C(10,3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120 ways
Key Takeaways
- Fundamental Counting Principle: Multiply the number of ways each event can occur
- Factorial n!: Product of integers from 1 to n (grows rapidly!)
- Permutations P(n,r): Use when ORDER MATTERS → n!/(n−r)!
- Combinations C(n,r): Use when ORDER DOESN'T MATTER → n!/(r!(n−r)!)
- Key question: "Does swapping items create a different outcome?" Yes = permutation, No = combination
- C(n,r) = P(n,r) / r! (combinations remove duplicate orderings)
What's Next?
You now have powerful tools for counting outcomes! In Lesson 3: Conditional Probability & Independence, you'll learn how probabilities change when you have partial information.
Coming up: P(A|B), multiplication rule, independent vs. dependent events