Probability & Counting · Grades 10–12
Permutations vs Combinations: How to Choose the Right One
The only question that matters: does order matter?
Gold, silver and bronze for three runners is an ordered result — swapping two runners changes the outcome. A three-person committee is not — the same three people form the same committee in any order.
Permutations (nPr)
Multiply the first r terms of n × (n−1) × (n−2) …. Ten runners competing for three medals can finish in 720 ways.
10P3
- 10P3 = 720
Step-by-step working 2 steps
- 1Order matters, no repeats
nPr = n! / (n−r)! = n × (n−1) × … (r factors)
Use permutations when the order of selection changes the outcome (race places, passwords without repeats).
- 2Multiply the first r terms
10 × 9 × 8 = 720
Combinations (nCr)
Divide nPr by r!, because each group of r items has r! different orderings that we no longer want to count separately. Choosing 3 from 10 gives 120 groups; choosing a five-card hand from a standard deck gives 2,598,960.
10C3
- 10C3 = 120
Step-by-step working 3 steps
- 1Order does not matter, no repeats
nCr = n! / (r!(n−r)!) = nPr / r!
Use combinations when you only care which items are chosen (committees, lottery picks).
- 2Compute nPr and r!
10P3 = 720 3! = 6
- 3Divide
720 ÷ 6 = 120
When items can repeat
A four-digit PIN allows repeated digits and order matters, so there are n^r = 10,000 possibilities.
4-digit PIN
- 10^4 = 10,000
Step-by-step working 1 steps
- 1Order matters, repeats allowed
n^r = 10^4 = 10,000
Each of the r positions can be filled in n ways, independently.
Common mistakes
Reaching for a formula before reading the question
Decide first whether order matters and whether repeats are allowed.
Thinking nCr can exceed nPr
It never does: nCr = nPr ÷ r!, so it is the same or smaller.
Forgetting repetition cases
PINs, dice sequences and passwords usually allow repeats.
Frequently asked questions
Is a combination lock a combination or a permutation?
Mathematically it is a permutation, because the order of the digits matters.
How do I get probability from a count?
Divide the number of favourable selections by the total number of equally likely selections, using the same counting rule for both.
