N5: Systematic Listing
Systematic listing strategies including the product rule for counting
📋 Key Concepts
Systematic Listing: A method of listing all possible outcomes or combinations in an organised way so none are missed or repeated.
Product Rule: If there are m ways to do one thing and n ways to do another, there are m × n ways to do both together.
Total combinations = choices for first item × choices for second item × ...
📝 Basic Listing
Method: Work through possibilities in order - fix one element and vary the others systematically.
Example 1
How many 2-digit numbers can be made using the digits 2, 3, 5?
Solution:
Starting with 2: 22, 23, 25
Starting with 3: 32, 33, 35
Starting with 5: 52, 53, 55
Total: 9 numbers
Using product rule: 3 × 3 = 9 ✓
📝 The Product Rule
Example 2
A menu has 3 starters, 4 main courses and 2 desserts. How many different 3-course meals can be chosen?
Solution:
Product rule: 3 × 4 × 2 = 24
Answer: 24 different meals
Example 3
How many different outcomes are there when rolling two dice?
Solution:
Each die has 6 possible outcomes
Product rule: 6 × 6 = 36
Answer: 36 possible outcomes
📝 Arrangements
Arrangements: When arranging items in order, the number of choices decreases by 1 each time.
Example 4
How many ways can 4 books be arranged on a shelf?
Solution:
First position: 4 choices
Second position: 3 choices
Third position: 2 choices
Fourth position: 1 choice
Total: 4 × 3 × 2 × 1 = 24 ways
📝 With and Without Repetition
With repetition: Each choice is independent - use simple product rule.
Without repetition: Each choice affects the next - multiply by decreasing values.
Without repetition: Each choice affects the next - multiply by decreasing values.
Example 5
How many 3-letter codes can be made from A, B, C, D:
(a) with repetition allowed?
(b) without repetition?
Solution:
(a) With repetition: 4 × 4 × 4 = 64 codes
(b) Without repetition: 4 × 3 × 2 = 24 codes
📝 Multiple Conditions
Example 6
A PIN code is 4 digits. How many possible codes if:
(a) all digits can be any number 0-9?
(b) the first digit cannot be 0?
Solution:
(a) 10 × 10 × 10 × 10 = 10,000
(b) 9 × 10 × 10 × 10 = 9,000
🧮 Worked Examples — Exam Style
Counting questions reward SYSTEMATIC listing (so nothing is missed or doubled) or the product rule (multiply choices at each stage).
A menu has 4 starters and 6 mains. How many different two-course meals?
Product rule: 4 × 6 = 24 meals
B1 for 24. Each starter pairs with each main — that is WHY you multiply.
List systematically all the ways to choose 2 letters from A, B, C (order doesn't matter).
AB, AC, BC — fix the first letter and pair it with each later one.
Listing AB, BA, AC, CA… would double-count: order doesn't matter here.
M1 for a systematic list, A1 for exactly 3. 'Order doesn't matter' halves the count versus arrangements.
Exam wording to expect: "How many different ways …", "List all the possible …", "How many combinations …".
❓ Practice Questions
Q1: A coin is flipped and a die is rolled. How many possible outcomes?
Q2: How many ways can 3 people line up for a photo?
Q3: A cafe offers 5 sandwiches, 3 drinks and 2 snacks. How many meal combinations?
Q4: List all possible outcomes when spinning a 3-sided spinner twice.
Q5: How many 2-digit numbers can be formed from digits 1, 2, 3, 4 without repetition?
Q6: How many ways can 4 friends arrange themselves in a row for a photo?
Q7: A passcode uses 2 letters (A–D, no repeats) then 1 digit (0–9). How many codes?
✅ Answers
- 2 × 6 = 12 outcomes
- 3 × 2 × 1 = 6 ways
- 5 × 3 × 2 = 30 combinations
- (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) = 9 outcomes
- 4 × 3 = 12 numbers
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. 2 × 6 = 12 outcomes
Q2. 3! = 3 × 2 × 1 = 6 ways
Q3. 5 × 3 × 2 = 30 combinations
Q4. Fix spinner 1 and vary spinner 2: (1,1), (1,2), (1,3), (2,1)… — 9 outcomes (3 × 3)
Q5. 4 choices for the first digit × 3 remaining = 12 numbers
Q6. 4 × 3 × 2 × 1 = 24 ways
Q7. 4 × 3 × 10 = 120 codes
📌 Key Facts to Memorise
- Product rule: multiply the number of choices at each independent stage
- Arrangements (order matters): n × (n−1) × … down to the slots filled
- Selections (order irrelevant): count pairs/groups once — AB and BA are the SAME
- Systematic listing: fix the first item, run through the rest, then move on
- With repetition allowed: the same number of choices at every stage
🗺️ In the Exam
Listing/counting: 1–3 marks on every paper, often as the setup for a probability question (count outcomes to get denominators).
Links to: possibility spaces (P7), tree diagrams (P8), probability (P3).
⚡ Targeted Drills
Cover the answers, work each one, then mark yourself. If any takes over a minute, revisit the matching section above.
- 3 shirts × 4 ties → 12 outfits · 5 × 5 grid paths (right/down only) → listing, not product rule
- Arrangements of 4 books → 4 × 3 × 2 × 1 = 24
- Pairs from 5 people → 5 × 4 ÷ 2 = 10 (divide by 2 — AB = BA)
- 3-digit codes from {1,2,3,4}, repetition allowed → 4³ = 64
- Spinner 1–3 twice, total 4 → (1,3), (2,2), (3,1) → 3 of 9
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always check if repetition is allowed or not
- Draw a table or tree diagram if the question asks to list all outcomes
- For arrangements without repetition, think: "How many choices for the first position? Second?"
- Check your answer is reasonable - if arranging n items, the answer should be less than n^n
- Read carefully - "at least" or "exactly" change the calculation
🧠 Problem-Solving Strategies
Problem-Solving
When listing combinations, always identify whether repetition is allowed. Use the product rule for counting: multiply the number of choices at each stage. When order matters (arrangements), choices decrease. When order doesn't matter (combinations), be careful not to overcount. Drawing a table or tree diagram helps for small cases to verify your product rule answer.
Multi-Step Problem
A password is 3 characters long. Each character is a letter from A to E or a digit from 1 to 4. The first character must be a letter and the last must be a digit. No character can be repeated. How many possible passwords are there?
Solution: First character: 5 letters. Last character: 4 digits. Middle: remaining characters = 5 + 4 − 1 = 8 (no repetition). Total = 5 × 8 × 4 = 160 passwords.
⚠️ Common Errors
Watch Out!
1. Wrong: Arranging 3 books has 3 × 3 × 3 = 27 ways (with repetition) Correct: Without repetition: 3 × 2 × 1 = 6 ways
2. Wrong: A 4-digit PIN with no restrictions has 9 × 10 × 10 × 10 = 9000 codes Correct: The first digit CAN be 0, so 10 × 10 × 10 × 10 = 10,000 codes
3. Wrong: Choosing 2 items from {A,B,C} without order gives 6 ways Correct: AB, AC, BC = 3 ways (order doesn't matter so AB = BA)
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A restaurant offers 3 starters, 5 main courses and 2 desserts. (a) How many different 3-course meals can be chosen? (b) The restaurant adds a new dessert. How many additional meal combinations does this create? (c) A customer who doesn't eat fish can choose from 2 starters, 4 mains and all desserts. What fraction of all possible meals can this customer choose?
(a) 3 × 5 × 2 = 30 meals.
(b) New desserts = 3. New total = 3 × 5 × 3 = 45. Additional = 45 − 30 = 15 meals.
(c) Customer's meals = 2 × 4 × 3 = 24. Fraction = 24/45 = 8/15.
Mark scheme: 2 marks for (a), 2 marks for (b) with clear working, 2 marks for (c) including simplified fraction
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A school trip needs groups of students. There are 30 students and 6 adult supervisors.
(a) Each group needs at least 1 adult. What is the maximum number of groups they can form?
(b) If they form 3 equal-sized groups, how many students are in each group?
(c) A student says "There are more ways to arrange the adults into 3 groups than the students because there are fewer adults." Is this correct? Explain your reasoning.
Answers: (a) 6 groups (one adult per group). (b) 30 ÷ 3 = 10 students per group. (c) No — arranging fewer items into groups gives fewer ways, not more. 6 adults into 3 groups of 2 has fewer arrangements than 30 students into 3 groups of 10. Fewer items means fewer possible arrangements overall.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Mathematics exams?
Get the best revision books and guides to boost your grades.