Quick answer: Choosing 3 of 10 people gives C(10, 3) = 10! ÷ (3! × 7!) = 720 ÷ 6 = 120 combinations; if the order matters (1st, 2nd, 3rd place) it is P(10, 3) = 10 × 9 × 8 = 720.
How it's calculated
Permutations count ordered arrangements: nPr = n! ÷ (n − r)! = n × (n − 1) × … × (n − r + 1). Gold, silver and bronze among 10 runners is P(10, 3) = 720.
Combinations ignore the order: nCr = n! ÷ (r! × (n − r)!) = nPr ÷ r!. A 3-person committee from 10 people is C(10, 3) = 120, because each group of 3 can be ordered in 3! = 6 ways.
With repetition allowed, ordered choices number nʳ (a 4-digit PIN has 10⁴ = 10,000), and unordered choices number C(n + r − 1, r).
Results are exact whole numbers, however large. To turn a count into a chance, divide favourable by total outcomes in the probability calculator.
Worked examples
- C(10, 3) and P(10, 3): 120 and 720
- Poker hands, C(52, 5): 2,598,960
- Lotto 6 of 49, C(49, 6): 13,983,816
Frequently asked questions
How do I know if order matters?
Ask whether swapping two chosen items gives a different result. Seats, ranks and passwords: order matters (permutations). Teams, hands of cards and lottery tickets: it does not (combinations).
What is 0! ?
1, by definition. That is why C(n, 0) = C(n, n) = 1: there is one way to choose nothing or everything.
How large can n be?
Up to 1,000. Very long results are shown in scientific notation in the tiles.
How many ways can I pick 3 people out of 10?
120, if it doesn't matter who's picked first. If it does, like first, second and third place, it's 720.
How many 4-digit PINs are there?
10,000, from 0000 to 9999. Digits can repeat, so it's 10 × 10 × 10 × 10.
What are the odds of winning a 6-from-49 lottery?
About 1 in 14 million. There are 13,983,816 different sets of six numbers and only one takes the jackpot.