Systematic listing and the product rule for counting
Counting questions feel deceptively simple — until you miss an outcome and lose all the marks. The two tools you need are systematic listing (write outcomes in a fixed order so nothing is missed) and the product rule (multiply when choices are independent).
Systematic listing — write outcomes in order
The trick is to fix one variable at a time and cycle through the others. This guarantees you neither miss outcomes nor count duplicates.
Worked example: How many two-digit numbers can you make from the digits 2, 3, 5 if digits cannot repeat?
Fix the tens digit, then list the units in order:
- Tens 2: 23, 25
- Tens 3: 32, 35
- Tens 5: 52, 53
Total: 6 numbers.
For more variables, use a tree diagram or a table — they enforce the system on you.
The product rule
If a process has independent stages, multiply the number of choices at each stage.
Stage 1 has m outcomes, Stage 2 has n outcomes, …, Stage k has p outcomes → Total = m × n × … × p.
Worked example: a menu has 4 starters, 6 mains and 3 desserts. How many three-course meals? 4 × 6 × 3 = 72 meals.
With or without repetition
This is the key question to ask yourself.
With repetition (digits can repeat): each stage has the same number of choices.
- 4-digit codes from 0–9 (digits may repeat): 10 × 10 × 10 × 10 = 10,000.
Without repetition (each item used once): the number of choices reduces by 1 each stage.
- 3-letter words from {A, B, C, D, E} no repeats: 5 × 4 × 3 = 60.
For full permutations of n distinct items: n × (n−1) × (n−2) × … × 1 = n! ("n factorial").
Restrictions — handle them first
Always deal with restrictions BEFORE counting the rest, because they cut the choices for that stage.
Worked example: How many 3-digit numbers between 100 and 999 are odd?
- Hundreds: 9 choices (1–9, no leading zero).
- Tens: 10 choices (0–9).
- Units: 5 choices (must be odd: 1, 3, 5, 7, 9). Total = 9 × 10 × 5 = 450.
Choosing vs arranging — order matters?
If order matters (e.g. running positions in a race), count using the product rule directly. If order does NOT matter (e.g. choosing 3 friends from 10 to invite), divide by the number of ways to arrange the chosen set: 10 × 9 × 8 ÷ (3 × 2 × 1) = 120.
This is "n choose r", written ⁿCᵣ or C(n, r), and is examined at Higher tier.
⚠Common mistakes— Common mistakes (examiner traps)
- Adding instead of multiplying when stages are independent. "OR" suggests +, but only when outcomes are mutually exclusive within ONE stage. Sequential choices ("AND") need ×.
- Forgetting to reduce when items can't repeat. 5 × 5 × 5 vs 5 × 4 × 3 — read carefully.
- Allowing leading zeros when the question forbids them (e.g. counting 4-digit numbers ≥ 1000).
- Listing without a system, so duplicates appear and items are missed.
- Confusing "arrangements" with "selections" — divide by r! when order doesn't matter.
➜Try this— Quick check
A 3-digit code uses the digits 1, 2, 3, 4, 5 with no repeats and the code must be even. How many codes are possible?
Last digit must be 2 or 4 → 2 choices. First digit: 4 remaining → 4 choices. Middle digit: 3 remaining → 3 choices. Total = 4 × 3 × 2 = 24 codes.
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