GCSE Maths · Number

Product rule for counting GCSE Questions and Worked Answers

Multiply the number of choices at each stage when each earlier choice has the stated number of continuations. Split different cases and remove duplicate counts when order does not matter.

Higher · listing prerequisites6 worked examples10 original questions
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Start with the meaning

What you need to know about product rule for counting

Choose one of two shirts, red or blue, and then one of three hats, A, B or C. With the red shirt there are three outfits. With the blue shirt there are another three. Counting 3 + 3 gives six: this is why multiplying 2 by 3 works.

See the idea first

One complete outcome uses one choice from each stage

The table lists every outfit once. A product counts repeated equal-size groups. It does not give a probability by itself, and the choices do not need to be equally likely just to count them.

All six outfits
ShirtHat AHat BHat C
RedRed–ARed–BRed–C
BlueBlue–ABlue–BBlue–C
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Count a multi-stage choice

What the problem asks: Choose one of four sandwiches and one of six drinks. How many meal choices exist?

How to solve it: Every sandwich has six possible drinks, so there are 4 × 6 = 24 choices.

Count without repetition

What the problem asks: In how many ways can a captain and a different deputy be chosen from seven pupils?

How to solve it: There are seven captain choices, then six remaining pupils for deputy: 7 × 6 = 42. The roles are different, so swapping pupils gives a different outcome.

Count unordered pairs

What the problem asks: How many different two-person teams can be chosen from seven pupils?

How to solve it: 7 × 6 counts each team twice, once in each order. Divide by two to get 21. No captain or deputy is being assigned.

A reliable routine

Count outcomes built in stages

The product rule applies when every partial outcome has the same number of allowed next choices at that stage. The number may shrink after a choice. If it varies between cases, count each separate case and add.

  1. Define one complete outcome and whether order matters.
  2. Write the available choices at each stage, applying restrictions immediately.
  3. Multiply along one complete case.
  4. Add disjoint cases, or divide by an explained duplication factor when the same outcome has been counted repeatedly.

Check: State whether digits may repeat and whether an initial zero is allowed. A code can start with zero; an ordinary three-digit number cannot.

Fully worked

Product rule for counting GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

A menu

Higher only2 marks
Question

A café offers five soups and four rolls. Choose one soup and one roll. How many meals are possible?

Each soup has four roll choices.

5×4=205\times4=20

There are 20 meals.

Example 2

Three stages

Higher only2 marks
Question

There are three jackets, two trousers and four pairs of shoes. How many outfits use one of each?

3×2=63\times2=6

jacket–trouser pairs. Each pair has four shoe choices:

6×4=246\times4=24

outfits.

Example 3

Repeated digits allowed

Higher only2 marks
Question

How many four-digit PIN codes use digits 0–9 if repetition and a first digit of zero are allowed?

Each of the four positions has ten choices.

10×10×10×10=1000010\times10\times10\times10=10000

Tip: 0000 is a valid code under these rules.

Example 4

No repeated digits

Higher only3 marks
Question

How many three-digit numbers use digits 1, 2, 3, 4, 5 with no repeated digit?

There are five choices for the first digit, then four, then three.

5×4×3=605\times4\times3=60

The order changes the number.

Example 5

One match per pair

Higher only3 marks
Question

Nine teams each play every other team once. How many matches are there?

Choose a team and its opponent: 9×8=729\times8=72. This counts A–B and B–A separately, but they are one match.

72÷2=3672\div2=36

There are 36 matches.

Example 6

Split a restriction into cases

Higher only4 marks
Question

How many two-digit even numbers use digits 0, 1, 2, 3, 4 without repetition?

If the units digit is 0, the tens digit has 4 choices. If it is 2, the tens digit can be 1, 3 or 4: 3 choices. If it is 4, there are another 3.

4+3+3=104+3+3=10

These cases are separate; an initial zero is not allowed.

10 original questions · total 25 marks

Product rule for counting GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 30 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Two choices

Higher only2 marks

Choose one of six cards and one of three envelopes. Count the combinations.

Show worked answer

Each card can use three envelopes:

6×3=186\times3=18
2

Three choices

Higher only2 marks

A badge has one of four shapes, five colours and two sizes. How many badges are possible?

Show worked answer
4×5×2=404\times5\times2=40

Each choice can be followed by every option at the next stage.

3

Named roles

Higher only2 marks

In how many ways can a chair and a different secretary be chosen from eight people?

Show worked answer

There are 8 choices then 7:

8×7=568\times7=56

Swapping the roles creates a different assignment.

4

A pair

Higher only3 marks

How many different two-person teams, with no distinct roles, can be chosen from eight people?

Show worked answer

Ordered selection gives 8×7=568\times7=56. Each pair appears in both orders:

56÷2=2856\div2=28
5

A letter and digits

Higher only2 marks

A code uses one of 26 letters followed by two digits 0–9. Digits may repeat. Count the codes.

Show worked answer
26×10×10=260026\times10\times10=2600

The letter stage and each digit stage have the stated choices.

6

No leading zero

Higher only3 marks

Count three-digit numbers using 0–9 with repetition allowed.

Show worked answer

The first digit has 9 choices, 1–9. The other positions each have 10.

9×10×10=9009\times10\times10=900
7

No repetition or leading zero

Higher only3 marks

Count three-digit numbers using 0–9 without repeated digits.

Show worked answer

The first digit has 9 choices. Nine digits remain, including zero, then eight.

9×9×8=6489\times9\times8=648
8

Routes

Higher only2 marks

There are four routes from A to B and three from B to C. How many journeys go from A to C through B, with any route combination allowed?

Show worked answer

Every first route can be followed by three second routes.

4×3=124\times3=12
9

Separate menu cases

Higher only3 marks

A deal is either one of three soups with one of two breads, or one of four salads with one of three dressings. Count the distinct deals.

Show worked answer

Soup deals: 3×2=63\times2=6. Salad deals: 4×3=124\times3=12. The categories do not overlap:

6+12=186+12=18
10

At least one occurrence

Higher only3 marks

How many three-digit PIN codes contain at least one 7? Digits 0–9 may repeat and zero may come first.

Show worked answer

Count all codes, then remove those with no 7.

103=100010^3=1000 93=7299^3=729 1000729=2711000-729=271

The complement avoids double-counting codes containing several 7s.

Examiner-style feedback

Common product rule for counting mistakes

Adding different stages

Adding counts soups or rolls separately; multiplying counts soup-and-roll combinations.

Missing a restriction

An already used item is unavailable when repetition is forbidden.

Dividing by two automatically

Divide only if every counted outcome appears exactly twice. Named roles and ordered codes usually do not.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. Define an outcome.
  2. Decide whether order matters.
  3. Count available choices at each stage.
  4. Explain any addition or division.
Quick answers

Product rule for counting FAQ

Is this probability?

It is counting. Counts can later be used for probability when the relevant outcomes are equally likely.

Must every stage have the same number of choices?

No. The number can change by stage, for example 5 then 4 then 3. What matters is that each partial outcome at a stage has that stated number of continuations.

Build connected skills

What to revise next

Content standards

Curriculum and rights review

N5: systematic-listing prerequisite followed by Higher-tier product-rule applications. No A-level factorial notation is required. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references