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Year 11 Methods (Unit 1 & 2) Counting Methods And Binomial Expansions

Addition And Multiplication Principles

20 practice questions 1 video lesson Theory + worked examples

Learn the addition and multiplication principles for Queensland Year 11 Mathematical Methods (QCAA). These counting rules tell you how many ways a task can be done: multiply choices made stage by stage, and add counts across separate cases.

You will learn to count tasks built in steps, split a problem into separate cases, and count with restrictions such as no repeats — supporting counting skills that lead into combinations and probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the multiplication principle counts outcomes built in independent stages (multiply the choices), while the addition principle counts outcomes that fall into separate, mutually exclusive cases (add the counts). This page shows how to choose and combine these counting methods for menus, codes, PINs and number plates, including counting with a restriction.

The multiplication principle applies when a task is completed in a sequence of independent stages or positions. If the first stage can happen in \(n_1\) ways, the second in \(n_2\) ways, and so on, the total number of outcomes is the product \(n_1\times n_2\times\cdots\). Think of it as filling separate boxes and multiplying.

The addition principle applies when the outcomes split into mutually exclusive cases that cannot happen together. If the cases have \(n_1,\,n_2,\dots\) outcomes, the total is the sum \(n_1+n_2+\cdots\).

A restriction such as no repetition means each choice reduces the options for the next stage, giving decreasing factors (for example \(8\times7\times6\)).

Multiply within a stage-by-stage task; add across separate cases. The word AND usually signals multiply; the word OR usually signals add.
Multiplication principle box diagramThree mains times four drinks fills two stages to give twelve meals. 3 mains × 4 drinks = 12
Multiplication principle: \(3\) mains \(\times\ 4\) drinks fills two stages to give \(12\) meals.
Two-stage counting treeA tree with two first-stage branches, each splitting into three, giving six paths. 6 paths
A tree: \(2\) first choices, each with \(3\) second choices, gives \(2\times3=6\) paths.

Multiplication principle (independent stages):

\[N=n_1\times n_2\times\cdots\times n_k\]
N=n1×n2××nk

Addition principle (mutually exclusive cases):

\[N=n_1+n_2+\cdots+n_k\]
N=n1+n2++nk
No repetition: the options fall by one each stage, so a code of \(r\) distinct items from \(n\) is \(n(n-1)(n-2)\cdots\) with \(r\) factors.

How to count outcomes

  1. Split: decide whether the task is one job done in stages (AND, multiply) or a choice between separate cases (OR, add).
  2. Count each part: draw a box for each stage and write how many options it has, applying any restriction (a fixed position, or no repetition).
  3. Combine: multiply the boxes within a stage-by-stage task, then add the totals of separate cases to get the final count.
Example 1 — Multiplication principle
A cafe offers \(3\) mains and \(4\) drinks. How many different meals of one main and one drink are possible?
Solution

Stages — a meal is one main AND one drink, so multiply the two stages:

\(N\)\(=\)\(3\times 4\)

Multiply:

\(N\)\(=\)\(12\)

There are \(12\) possible meals.

N=3×4=12
Example 2 — Addition principle
To get to school Mia can take one of \(3\) buses or one of \(2\) trains. In how many ways can she travel?
Solution

Cases — a bus OR a train are mutually exclusive, so add the two cases:

\(N\)\(=\)\(3+2\)

Add:

\(N\)\(=\)\(5\)

There are \(5\) ways to travel.

N=3+2=5
Example 3 — Counting with no repetition
A \(4\)-letter code is made from the \(8\) letters \(A\) to \(H\) with no letter repeated. How many codes are possible?
Solution

Four positions — each filled letter removes one option from the next:

\(N\)\(=\)\(8\times 7\times 6\times 5\)

Multiply left to right:

\(N\)\(=\)\(56\times 6\times 5\)
\(=\)\(336\times 5\)
\(=\)\(1680\)

There are \(1680\) codes.

N=8×7×6×5=1680
Example 4 — A restriction (even numbers)
How many \(3\)-digit numbers can be formed using the digits \(1,2,3,4,5\) with no digit repeated, if the number must be even?
Solution

Start with the restricted position — even means the last digit is \(2\) or \(4\):

\(\text{units}\)\(=\)\(2\text{ ways}\)

Then fill the remaining two positions from the \(4\) digits left:

\(\text{hundreds, tens}\)\(=\)\(4\times 3\)

Multiply all three positions together:

\(N\)\(=\)\(4\times 3\times 2\)
\(=\)\(24\)

There are \(24\) such even numbers.

N=4×3×2=24

Common pitfalls

Adding when you should multiply. If one task is built in stages (choose a main AND a drink), multiply. Only add when the outcomes are separate cases (a bus OR a train).
Forgetting the numbers fall with no repetition. After you use a letter it is gone, so the factors decrease \(8\times7\times6\), not \(8\times8\times8\).
Ignoring the restricted position first. Fill the position with the restriction (even, or cannot be \(0\)) before the free positions, or you will overcount.

Frequently asked questions

When do I add and when do I multiply?

Multiply when one task is completed in stages (the word AND); add when the outcome falls into separate, mutually exclusive cases (the word OR).

What is the multiplication principle?

If a task has independent stages with \(n_1,n_2,\dots\) options, the total number of outcomes is the product \(n_1\times n_2\times\cdots\).

How does no repetition change the count?

Each choice removes one option from the next stage, so the factors decrease: a \(4\)-item code from \(8\) letters is \(8\times7\times6\times5\).

How do I handle a restriction like a leading digit that cannot be zero?

Fill the restricted position first (for example \(9\) choices for a non-zero first digit), then multiply by the choices for the remaining positions.

Can I use both principles in one question?

Yes. Multiply within each case to count its outcomes, then add the case totals together for the final answer.