When order matters: Here, order matters — like who finishes 1st, 2nd or 3rd, or who sits in which seat.
To count all the ways, fill the positions one at a time and multiply the choices.
IB-style question — distinct-digit numbers
How many 4-digit numbers can be formed using the digits 0–9 if no digit is repeated?
Step by step
- A box for each digit. The first box can't be 0 (else 0527 is just 527), so 9 choices: 1–9.
- Now 0 is allowed, but one digit is used: 9 left, then 8, then 7.
- Multiply the boxes.
Final answer
4536 numbers.
[Diagram: math-choice-boxes] - Available in full study mode
The key idea — choices run out: Each slot shows how many digits you can still use. Every digit you place is used up, so the next slot has one fewer choice. That's why the numbers count down: 9, 8, 7.
Apply it — a phone passcode
You set a 4-digit phone passcode using the digits 0–9 with no digit repeated.
How many passcodes are possible?
Step by step
- Same method — a slot for each digit. But a passcode CAN start with 0 (0527 is a fine passcode), so slot 1 has all 10 choices.
- Each digit used removes one choice for the next slot.
- Multiply.
Final answer
5040 passcodes (more than the 4536 numbers — because the passcode is allowed to start with 0).
Arranging everything: Picture n people lining up for a photo. The first spot has n people to choose from; once someone takes it, only n − 1 are left for the next spot, then n − 2 … down to 1 for the last spot.
Multiply the spots together to count every line-up.
IB-style question — a whole line
9 people stand in a line for a photograph.
In how many different orders can they stand?
Step by step
- 1st spot: any of the 9. Once that person stands there, they're used up — so only 8 are left for the 2nd spot, then 7, … down to 1 for the last spot.
- Multiplying every number down to 1 is exactly 9 factorial.
Final answer
362 880 orders.
IB-style question — finishing order
8 sprinters run a final with no tied finishes.
In how many different orders can they finish?
Step by step
- 1st place: any of the 8 runners. Once one finishes, only 7 are left for 2nd, then 6, … down to 1 for last.
- That product down to 1 is 8 factorial.
Final answer
40 320 orders.
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All of them, or only some?: All of them in a row → n! (last section).
Only some — say r out of n — in order → ⁿPᵣ, like picking a captain, then a vice-captain from a team.
- how many objects you can choose from
- how many you arrange, in order
IB-style question — awarding medals
8 sprinters run a final with no ties.
In how many ways can the gold, silver and bronze medals be awarded?
Step by step
- The 3 medals are different — 1st = gold, 2nd = silver, 3rd = bronze. Swapping them changes who wins what, so order matters: arrange 3 of the 8 → ⁸P₃.
- Gold: any of the 8. Silver: 7 runners left. Bronze: 6 left.
Final answer
336 ways.
IB-style question — choosing seats
5 students sit down in a row of 10 empty chairs (one student per chair).
In how many ways can they be seated?
Step by step
- The seats are different (seat 1 ≠ seat 2). So Ana in seat 1, Ben in seat 2 is a different seating from the swap — who sits where matters, so use P → ¹⁰P₅.
- Now just count, one student at a time. (The numbers drop only because a seat gets used up each time.) 1st student: 10 chairs; 2nd: 9 left; then 8, 7, 6.
Final answer
30 240 ways.
When would this be 'choose' instead?: If the question only asked which 5 chairs get used — not who sits in them — then swapping two people wouldn't matter, and you'd get fewer ways. That 'order doesn't matter' case is the next section.