Introduction to Combinatorics & Counting Principles
A restaurant has 3 entrees, 5 mains, and 2 desserts. How many different meals could you order? A PIN uses 4 digits — how many are possible? Combinatorics is the mathematics of counting systematically, and it underpins probability, cryptography, and everything in between. This lesson builds the foundational tool: the Fundamental Counting Principle.
A bike lock has 4 dials, each showing digits 0–9. Without calculating — how many different lock combinations do you think are possible? Write your gut estimate and your reasoning.
Combinatorics rests on two core rules: the Multiplication Principle (AND — do one thing and another) and the Addition Principle (OR — do one thing or another). This lesson focuses on multiplication; addition comes in later lessons.
Every multi-stage counting question reduces to one of two operations: multiply the options at each independent stage (AND logic), or add disjoint cases (OR logic). Identifying which applies is the first — and most important — step.
Key facts
- The Fundamental Counting Principle: $n_1 \times n_2 \times \cdots \times n_k$
- With repetition: $n^k$ arrangements for $k$ stages from $n$ options
- "And" means multiply; "or" means add (for disjoint cases)
Concepts
- Why multiplying stage-by-stage counts every possible combination
- The difference between arrangements with and without repetition
- Why the order in which you apply restrictions matters
Skills
- Identify the number of stages and options in a counting problem
- Apply the Fundamental Counting Principle to multi-stage problems
- Handle problems with and without repetition restrictions
Combinatorics is the branch of mathematics concerned with counting, arranging, and selecting objects. It answers questions like "How many ways can…?"
In the HSC, combinatorics appears in:
- Counting arrangements of people, letters, or objects
- Selecting committees or teams
- Expanding binomial expressions (Module 4 later lessons)
- Probability calculations (both Maths Advanced and Extension 1)
The core tool is the Fundamental Counting Principle: if a task can be broken into independent stages, the total count is found by multiplying the number of choices at each stage.
Combinatorics = counting arrangements and selections systematically; Fundamental Counting Principle: k stages with n_1, n_2, , n_k choices total = n_1 n_2 n_k
Pause — copy the combinatorics definition into your book: combinatorics counts systematic arrangements and selections; Fundamental Counting Principle: $k$ stages with $n_1, n_2, \ldots, n_k$ independent choices gives $n_1 \times n_2 \times \cdots \times n_k$ total outcomes.
Quick check: A café offers 4 types of coffee and 3 types of cake. How many coffee-and-cake combinations are possible?
We just saw that combinatorics counts arrangements and selections systematically. That raises a question: when there are multiple independent stages to a task, how do you count the total number of outcomes without listing them all? This card answers it → the Fundamental Counting Principle: multiply the number of choices at each stage, $n_1 \times n_2 \times \cdots \times n_k$.
If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
This principle extends to any number of stages: if a task consists of $k$ stages, with $n_1$ ways to complete stage 1, $n_2$ ways to complete stage 2, and so on, the total number of ways to complete the entire task is the product $n_1 \times n_2 \times \cdots \times n_k$.
Key distinction:
- With repetition: each stage is independent — the same option can be chosen more than once. For $k$ stages each with $n$ options: total $= n^k$.
- Without repetition: once an option is used, it cannot be reused. The number of options decreases by 1 at each stage.
With repetition: n options, k stages n^k total outcomes; Without repetition: n options for stage 1, n-1 for stage 2, n-2 for stage 3, etc.
Pause — copy the repetition/no-repetition comparison into your book: with repetition — $n$ options at each of $k$ stages gives $n^k$ outcomes; without repetition — $n \times (n-1) \times (n-2) \times \cdots$ with one fewer option at each stage.
Did you get this? True or false: a 3-digit number formed from digits 1–5 without repetition gives $5 \times 5 \times 5 = 125$ arrangements.
Worked examples · 3 in a row, reveal as you go
A restaurant offers 3 entrees, 5 mains, and 2 desserts. How many different three-course meals are possible?
How many three-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if: (a) digits can be repeated? (b) digits cannot be repeated?
A password is formed from 2 letters (A–Z) followed by 3 digits (0–9). How many passwords are possible if repetition is allowed?
Fill the gap: A 4-digit PIN using digits 0–9 with repetition allowed has possible PINs.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the number of 3-digit numbers that can be formed from digits 0–9 (no repetition, no leading zero) is $9 \times 9 \times 8 = 648$.
Activities · practice with the ideas
A car is available in 4 colours, 2 engine types, and 3 trim levels. How many different cars are possible?
How many 4-digit PINs can be made from digits 0–9 if no digit is repeated?
How many three-digit numbers greater than 500 can be formed from $\{1, 2, 3, 4, 5, 6, 7\}$ without repetition?
A student must choose 1 subject from each of: 3 Science electives, 4 Humanities electives, and 2 Creative Arts electives. How many subject combinations are possible?
Explain why the answer to a "with repetition" problem is always greater than or equal to the "without repetition" answer for the same $n$ and $k$.
Odd one out: Three of these four expressions correctly count the number of 2-digit codes from digits 1–4. Which one does NOT?
Earlier you estimated the number of combinations for a 4-dial bike lock (digits 0–9).
Each dial has 10 choices (0–9), and the dials operate independently (with repetition). By the Fundamental Counting Principle:
$$\text{Total} = 10 \times 10 \times 10 \times 10 = 10^4 = 10{,}000$$
There are exactly 10,000 combinations. If you estimated around this value, your intuition was good! If you underestimated, remember that multiplication grows very quickly — this is the power (and the danger) of combinatorics.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A password is formed from 2 letters followed by 3 digits. How many passwords are possible if repetition is allowed? (2 marks)
Q2. How many ways can 4 people be arranged in a line? (1 mark)
Q3. A code consists of 3 digits chosen from 0–9 without repetition. How many codes are possible? Justify why the answer differs from the "with repetition" case. (2 marks)
Comprehensive answers (click to reveal)
Activity 1: 1. $4 \times 2 \times 3 = 24$ · 2. $10 \times 9 \times 8 \times 7 = 5040$ · 3. Hundreds digit $\in \{5,6,7\}$ (3 choices); tens digit: 6 remaining choices; units digit: 5 remaining choices $\Rightarrow 3 \times 6 \times 5 = 90$ · 4. $3 \times 4 \times 2 = 24$ · 5. With repetition, each stage retains the full $n$ choices; without repetition each stage loses one option, so the product is always $\leq$ the with-repetition product.
Q1 (2 marks): $26 \times 26 \times 10 \times 10 \times 10 = 26^2 \times 10^3 = 676{,}000$ [2].
Q2 (1 mark): $4 \times 3 \times 2 \times 1 = 24$ ways [1].
Q3 (2 marks): Without repetition: $10 \times 9 \times 8 = 720$ [1]. With repetition: $10^3 = 1000$. They differ because without repetition each digit is removed from the pool once chosen, reducing the available choices at each subsequent stage [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering counting principle questions. A lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.