Build a count using product and sum rules
Describe what one outcome contains, multiply successive choices and add mutually exclusive cases.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.2. State whether order matters, objects are distinguishable, repetition is permitted and positions or rows are labelled. Count mutually exclusive cases before adding them; account for internal arrangements when using blocks. These lessons include seating in two or more rows. Arrangements in a circle are excluded from this syllabus section. Unless a question says otherwise, people and named roles are distinguishable, and selection is without replacement. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Multiply and add whole-number counts.
- List small sets of outcomes systematically.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which outcome needs one choice from each of two sets?
What you will learn
- Use the product rule when each outcome requires successive choices.
- Add disjoint alternatives without counting an outcome twice.
Define the complete outcome
An outfit consisting of one of three shirts and one of four pairs of trousers has 3 × 4 = 12 possibilities. The word ‘and’ means the outcome contains both choices. A tree makes this concrete: every shirt branch has four trouser branches. Multiplication is a counting rule here; it does not need a probability assumption about independence.
If later choices depend on earlier ones, count the available choices at each stage. A three-character code with first character A or B and last two characters different digits from 1, 2, 3 has 2 × 3 × 2 = 12 possibilities. The second digit has only two choices after the first has been used. Permitting digit repetition instead gives 2 × 3 × 3 = 18.
Add alternatives only when they are disjoint
A journey uses exactly one of six distinct bus services or four distinct train services. These ten named services form disjoint alternatives, so the count is 6 + 4 = 10. It is not 6 × 4: that would describe choosing a bus and a train. An outcome satisfying two overlapping descriptions must not be counted twice.
Before calculating, describe the object being counted and any restrictions. A tree or small list can reveal a missing branch or duplicate. If case sizes differ, count within each case and then add: one universal product may not apply. The same careful outcome definition will later support probability calculations.
Pause and explain
Two letters and three digits form one-letter, one-digit codes. How many?
Worked example
List codes whose first character is A or B and whose second character is 1, 2 or 3.
Show the worked solution
- For A, the codes are A1, A2 and A3.
- For B, they are B1, B2 and B3; these cases do not overlap.
- There are 2 × 3 = 6 codes, each with one letter and one digit.
Answer A1, A2, A3, B1, B2, B3: 6 codes.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
How many outfits use one of three shirts and one of four pairs of trousers?
Give me a hint
Each shirt can be paired with every pair of trousers.
Compare my reasoning
- Fix one shirt and list its four trouser choices.
- Repeat for all three distinct shirts; the groups do not overlap.
- Multiply 3 × 4 to obtain 12 complete outfits.
12 outfits.
Look for these in your work
- Count an outfit containing both items.
- Explain why each first choice has four continuations.
A journey uses exactly one of six named bus services or four named train services. All ten services are distinct. How many choices are there?
Give me a hint
The traveller chooses one service, not a bus–train pair.
Compare my reasoning
- There are six bus choices and four train choices.
- The alternatives are mutually exclusive for this single-service journey.
- Add 6 + 4 = 10 choices.
10 service choices.
Look for these in your work
- Use disjoint alternatives.
- Explain why multiplying would describe a different journey.
A code starts with A or B, followed by two different digits from 1, 2, 3. Count codes, then compare with permitting repeated digits.
Give me a hint
After the first digit, ask how many unused digits remain.
Compare my reasoning
- There are two first-letter choices and three first-digit choices.
- Without repeated digits, the final choice has size two: 2 × 3 × 2 = 12.
- With repeated digits permitted, it has size three: 2 × 3 × 3 = 18.
12 without digit repetition; 18 with repetition.
Look for these in your work
- Apply the restriction at the correct stage.
- Keep letter and digit positions distinct.
Common mistakes
- Define a complete outcome before choosing an operation.
- Reduce later choices when repetition is forbidden.
- Do not add overlapping cases without correcting the overlap.
What can you explain now?
Does multiplying choice counts require probabilistic independence?
Compare with the explanation
No; count the number of options available at each stage.
The product rule describes a sequence of choices with known branch counts.
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Make it stick
Describe one outcome; multiply successive choices, then add disjoint cases.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.