Learn counting methods and factorial notation. Math 30-1 Alberta curriculum.
Lesson 7.1 of Probability Distributions in Math 30-1 — Alberta curriculum lessons.
Definition. The Fundamental Counting Principle (FCP) is a way to figure out the total number of outcomes in a situation with multiple stages.
You calculate it by multiplying the number of possibilities at each stage.
The Key Idea. If you have to make a series of decisions one after another, the total number of ways to make all those decisions is found by multiplying the number of choices at each step.
Each decision is independent — what you choose at one step doesn't change how many options you have at the next step (unless the problem says otherwise).
Rule: Multiply when decisions happen together (AND). Add when you're counting separate cases (OR).
Example 1 — How many ways can the letters ABC be arranged?. Step 1: Understand what's being asked.
We have three letters: A, B, and C. We want to know how many different orders we can put them in. This is called an arrangement or a permutation.
Step 2: List them out to see the pattern.
ABC, BAC, CAB, ACB, BCA, CBA
Step 3: Count them.
There are 6 arrangements. We'll come back to why this is exactly 3 × 2 × 1 once we learn factorial notation in Tab 2. For now, listing works fine for small problems.
→ 6 ways
Example 2 — Coloured wooden toys. A toy manufacturer makes a wooden toy in three parts:
• Part 1 (top): red, white, or blue → 3 options
• Part 2 (middle): orange or black → 2 options
• Part 3 (bottom): yellow, green, pink, or purple → 4 options
How many different coloured toys can be produced?
Step 1: Identify the decisions.
We make 3 separate decisions — choosing a colour for the top, middle, and bottom. Each decision is completely independent.
Step 2: Count the options at each step.
arrayccccc & & & & [6pt] 3 & × & 2 & × & 4 array
These numbers come directly from the problem — the top piece has 3 colour options, the middle has 2, and the bottom has 4.
Step 3: Apply the FCP.
3 × 2 × 4 = 24 ways
Step 4: Verify with a smaller example.
If you only had the top and middle, you'd have 3 × 2 = 6 combinations. Each of those 6 combos can be paired with any of the 4 bottom colours, giving 6 × 4 = 24. ✓
→ 24 ways
Example 3 — Arranging letters ABCDEF. How many ways can the letters ABCDEF be arranged?
Step 1: Identify the decisions.
We have 6 letters and 6 positions to fill, one position at a time, left to right.
Step 2: Count the choices at each position.
arrayccccccccccc & & & & & & & & & & [6pt] 6 & × & 5 & × & 4 & × & 3 & × & 2 & × & 1 array
Position 1 can be any of the 6 letters. Once we place one, it's used up — so Position 2 only has 5 remaining letters to choose from, then 4 for Position 3, then 3, 2, and finally just 1 letter left for the last spot. Each blank gets one fewer choice because we can't reuse a letter.
Step 3: Multiply.
6 × 5 × 4 × 3 × 2 × 1 = 720 ways
Key insight: Each time we fill a position, we have one fewer choice because we can't reuse a letter. This is called arranging without repetition.
→ 720 ways
Example 4 — Car versions. A new car is available in 3 models, 6 colours, 2 transmissions, and 4 option packages. How many versions can be created?
Step 1: Identify the decisions.
We are choosing one option from each of four independent categories.
Step 2: Count options at each step.
arrayccccccc & & & & & & [6pt] 3 & × & 6 & × & 2 & × & 4 array
These numbers come directly from the problem — 3 model options, 6 colour options, 2 transmission options, and 4 package options.
Step 3: Apply the FCP.
3 × 6 × 2 × 4 = 144 versions
→ 144 versions
Example 5 — Quiz answers (repetition allowed). A math quiz has 8 multiple choice questions, each with choices A, B, C, or D. How many different sets of answers are possible?
Step 1: Identify the decisions.
We are choosing an answer for each of 8 questions. Each question is an independent decision.
Step 2: Count the choices at each step — 4 options per question, the same for all 8.
arrayccccccccccccccc & & & & & & & & & & & & & & [6pt] 4 & × & 4 & × & 4 & × & 4 & × & 4 & × & 4 & × & 4 & × & 4 array
Every question always has exactly 4 choices: A, B, C, or D. Because repetition is allowed — you can choose A on question 1 and A again on question 2 — the count never decreases. All 8 blanks stay at 4.
Step 3: Apply the FCP.
4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4^8 = 65,536
Key insight: Unlike Example 3, the choices here do NOT decrease. That's because repetition is allowed — you can answer A on question 1 and also A on question 2. When repetition is allowed, the count stays the same at every step.
→ 65,536
Example 6 — Arranging 5 paintings on a wall. In how many ways can 5 paintings be arranged on a wall?
Step 1: Identify the decisions.
There are 5 spots on the wall. We decide which painting goes in each spot, one at a time.
Step 2: Count the choices at each spot.
arrayccccccccc & & & & & & & & [6pt] 5 & × & 4 & × & 3 & × & 2 & × & 1 array
Spot 1 can hold any of the 5 paintings. Once a painting is placed, it can't go anywhere else — so Spot 2 has only 4 paintings left to choose from, then 3, 2, and finally just 1.
Step 3: Multiply.
5 × 4 × 3 × 2 × 1 = 120 ways
→ 120 ways
Example 7 — Arranging letters of BOLAGNY. How many ways can the letters of BOLAGNY be arranged?
Step 1: Check for repeated letters.
B, O, L, A, G, N, Y — all 7 letters are different. No repeats.
Step 2: Identify the decisions.
7 letters, 7 positions. Fill each position one at a time.
Step 3: Count choices at each position.
arrayccccccccccccc & & & & & & & & & & & & [6pt] 7 & × & 6 & × & 5 & × & 4 & × & 3 & × & 2 & × & 1 array
Position 1 can be any of the 7 letters. Once used, that letter is gone — so Position 2 has 6 remaining choices, then 5, 4, 3, 2, and just 1 letter left for the final spot.
Step 4: Multiply.
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways
→ 5040 ways
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