Probability and statistics · GCSE Maths
Product rule for counting
GCSE Maths product rule: multiply choices for independent stages — menus, outfits, PINs and multi-branch tree counts when order matters or not.
If you can do A in m ways and B in n ways independently, together there are m × n ways. Multiply along the stages.
The important bits
What you need to know
- 1
The product rule: if one choice can be made in m ways and a second in n ways (independently), there are m × n combined outcomes.
- 2
Extended: for k independent stages with n₁, n₂, …, nₖ choices, total = n₁ × n₂ × … × nₖ.
- 3
Example: 4 shirts and 3 trousers gives 4 × 3 = 12 outfits. Each shirt pairs with every trouser.
- 4
Menus: 3 starters, 5 mains, 2 desserts → 3 × 5 × 2 = 30 three-course meals if any combination is allowed.
- 5
PINs: 4 digits 0–9 with repetition allowed → 10 × 10 × 10 × 10 = 10 000 codes. First digit 10 choices, same for each position.
- 6
Without repetition: first digit 10, second 9, third 8 → 10 × 9 × 8. Order matters — 123 is different from 321.
- 7
Distinguish product rule (AND across stages) from addition rule (OR between mutually exclusive options).
- 8
Tree diagrams for counting list branches; multiply along a path. The product rule is the tree without drawing every branch.
Quotations worth analysing
Short evidence. Real method.
“m × n ways”
Multiply when stages are independent and you need one choice from each. Adding m + n is wrong unless the stages are OR alternatives.
“Multiply along the branches”
Counting and probability share the structure. For counting, each branch is a choice, not a probability — but the multiply rule is identical.
“Order matters for PINs and permutations”
123 and 321 count as different when order matters. For combinations (order free), divide by repeats — product rule alone overcounts.
Go deeper
Structured counting problems
A meal deal: 4 sandwiches, 3 drinks, 2 snacks. Total meals = 4 × 3 × 2 = 24. If vegetarian sandwiches are only 2 of the 4 but drinks and snacks unchanged: 2 × 3 × 2 = 12 veg meals. Restrict one stage, multiply the rest. Licence plates: 2 letters (26 each) then 3 digits (10 each): 26 × 26 × 10 × 10 × 10 = 676 000 if repetition allowed. If no repeated letters: 26 × 25 × 10³. Write stages as a product line so the examiner sees the structure.
Go deeper
Product rule versus addition
You can travel by bus (3 routes) or train (2 routes) — mutually exclusive, total 3 + 2 = 5 ways. You choose a bus route then a train ticket for a return involving both — that is not 3 + 2 unless the question asks OR. “Bus or train” → add. “Shirt and trousers” → multiply. “Starter and main” → multiply. Venn problems use add with care for overlap. Product rule never adds across stages that must all happen.
Go deeper
When repetition is forbidden
Three-digit codes from 1–9 without repetition: first digit 9, second 8, third 7 → 9 × 8 × 7 = 504. Each stage shrinks. If 0 is allowed but not as first digit: 9 × 9 × 8 for three-digit numbers. Identical items in a row (AAB) need division after the product rule to remove permutations of identical objects — that is beyond pure product rule but appears as “how many arrangements of the word…” on some Higher papers. Product rule gives 3! = 6 for ABC; for AAB divide by 2! for the repeated A.
See the idea in action
A café offers 5 sandwiches, 4 drinks and 3 cakes. How many lunch combinations (one of each) are possible? Independent stages: sandwich × drink × cake. Total = 5 × 4 × 3 = 60 combinations. If only 2 sandwiches are vegan and a vegan customer must pick vegan sandwich but any drink and cake: 2 × 4 × 3 = 24.
Exam technique
Turn knowledge into marks
List the stages as a product before calculating. “And” across stages → multiply. “Or” between exclusive options → add.
Common mistakes
Do not give these marks away
- 01
Adding choices from independent stages (5 + 4 + 3) instead of multiplying.
- 02
Ignoring restrictions on one stage (e.g. no repeated digits) and using 10 each time.
- 03
Using the product rule when options are mutually exclusive alternatives (should add).
3 hats and 6 scarves. How many hat–scarf pairs?
A18
B9
C3
D36
Show the answer
18. Each hat with each scarf: 3 × 6 = 18. 9 is 3 + 6. 36 is 6 × 6 or doubling error.
Quick questions
If this is the bit you searched
When do I multiply and when do I add?
Multiply for independent stages that all happen (AND). Add for mutually exclusive options where only one happens (OR).
Is the product rule the same as on probability trees?
Same structure: multiply along a path. For counting, branches are numbers of choices, not probabilities.
What if repetition is not allowed?
Reduce the number of choices at each stage after previous picks. First 10, second 9, third 8, etc.
Does order always matter?
For PINs and positions, yes. For “pick a team of 3 from 10” order does not matter — combinations, not pure product rule.