Basic counting methods
Master basic counting methods for Year 11 Specialist Mathematics in Queensland (QCAA). These are the two rules that start combinatorics: the multiplication principle for choices made in stages, and the addition principle for a single choice from separate groups.
You will learn to decide when to multiply and when to add, count arrangements with and without repeats, and combine stages and cases in one problem — the counting foundation for permutations, combinations and probability later in the course.
Theory
Basic counting methods are the two rules behind all of combinatorics in Year 11 Specialist Mathematics (QCAA, Queensland). The multiplication principle counts choices made in stages, and the addition principle counts a single choice from separate groups. This page shows when to multiply and when to add, with worked examples.
In a counting problem you build an outcome by making one or more choices. Two rules cover almost every case.
The multiplication principle: if a task is completed in stages, and stage 1 can be done in \(m\) ways, stage 2 in \(n\) ways, and so on, then the whole task can be done in \(m\times n\times\cdots\) ways. Use it when choices are made one after another ("and").
The addition principle: if the outcome is a single choice from two or more separate groups of sizes \(m\) and \(n\), the number of choices is \(m+n\). Use it for a choice from one group or another.
The deciding question is always "and" or "or": stages joined by "and" are multiplied; separate groups joined by "or" are added.
For a task completed in \(k\) independent stages, with \(n_1,n_2,\dots,n_k\) choices at each stage:
For a single choice from \(k\) separate groups:
How to count a task
- Break the task into stages, or into separate cases.
- Decide "and" or "or": stages done together use the multiplication principle; a single choice from separate groups uses the addition principle.
- Count the options at each stage or group, watching whether repeats are allowed.
- Combine: multiply across stages, add across separate cases.
Count the choices at each stage, then multiply (the two stages happen together):
| \(\text{shirts}\) | \(=\) | \(4\) |
| \(\text{shorts}\) | \(=\) | \(3\) |
| \(\text{outfits}\) | \(=\) | \(4 \times 3\) |
| \(=\) | \(12\) |
\(12\) outfits.
Each position is an independent stage; with repeats allowed every position keeps its full count:
| \(\text{letters}\) | \(=\) | \(26 \times 26 \times 26\) |
| \(\text{digits}\) | \(=\) | \(10 \times 10 \times 10\) |
| \(N\) | \(=\) | \(26^3 \times 10^3\) |
| \(=\) | \(17\,576\,000\) |
\(17\,576\,000\) plates.
Fill each position in turn; with no repeats the count falls by one each stage:
| \(\text{hundreds}\) | \(=\) | \(5\) |
| \(\text{tens}\) | \(=\) | \(4\) |
| \(\text{units}\) | \(=\) | \(3\) |
| \(N\) | \(=\) | \(5 \times 4 \times 3\) |
| \(=\) | \(60\) |
\(60\) numbers.
Count the via-\(B\) route by multiplying its two legs, then add the separate direct case:
| \(\text{via } B\) | \(=\) | \(3 \times 4\) |
| \(=\) | \(12\) | |
| \(\text{direct}\) | \(=\) | \(2\) |
| \(\text{total}\) | \(=\) | \(12 + 2\) |
| \(=\) | \(14\) |
\(14\) ways.
Common pitfalls
Frequently asked questions
When do you multiply and when do you add in counting?
Multiply when a task is done in stages, one after another ("and"). Add when the outcome is a single choice from separate groups ("or").
What is the multiplication principle?
If a task is completed in stages with \(m\), \(n\), \(\dots\) choices at each stage, the number of ways to complete the whole task is \(m\times n\times\cdots\).
What is the addition principle?
If you make a single choice from separate groups of sizes \(m\) and \(n\), the number of choices is \(m+n\).
How do you count arrangements with no repeats?
Fill each position in turn; because a used item cannot be reused, the count falls by one each stage, for example \(5\times4\times3\).
Do the letters and numbers on a number plate multiply?
Yes. Each position is an independent stage, so multiply the number of options for every position; if repeats are allowed each position keeps its full count.