Overcounting & Restrictions
The hardest counting questions add a twist that makes the naïve product wrong, and there are three reliable repair tools. First, the complement: to count arrangements with "at least one" of something, count the total and subtract the arrangements with none — usually far faster than adding cases. Second, glue and treat as a block: when items must sit together, bind them into one unit, arrange the units, then multiply by the internal arrangements of the block. Third, handle the restricted slot first: if a code cannot start with 0, choose the first position from its allowed options before filling the rest. The deepest idea is overcounting — whenever your method counts each valid outcome more than once, divide by the size of that repeat. This is exactly why combinations divide permutations by r!, and why a round table divides by n. The moment a scenario feels like it double-counts, ask "by what factor?" and divide.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core counting
| Counting principle | if step 1 has m ways and step 2 has n ways, together m × n ways |
|---|---|
| Factorial | n! = n × (n−1) × … × 2 × 1; 0! = 1 |
| Permutations (order matters) | nPr = n! / (n − r)! |
| Combinations (order does not) | nCr = n! / [r! (n − r)!] |
Handy identities
| Symmetry | nCr = nC(n − r) |
|---|---|
| Choose all or none | nC0 = nCn = 1 |
| Circular arrangements | (n − 1)! ways to seat n around a round table |
| Permutations with repeats | n! / (p! q! …) for repeated items |