Combination Calculator C(n, r)
Work out how many ways there are to choose r of n items when order does not matter: with or without repetition, exact up to 10,000 items, with the ordered count and the chance of one particular choice.
- Number of digits3
- Ordered choices P(n, r) = r! × C(n, r)720
- Chance of one particular choice0.833%
Ways to choose k of 10
How it is calculated
Combinations count choices where order does not matter: {A, B, C} and {C, B, A} are the same choice. When order matters (race places, passwords) use permutations; the two are linked by P(n, r) = r! × C(n, r). The values C(n, r) form row n of Pascal’s triangle and are the coefficients of (a + b)ⁿ.
C(n, r) = n! ÷ (r! × (n − r)!) · With repetition: C(n + r − 1, r)n: number of items · r: number chosen. Without repetition r ≤ n; with repetition an item may be chosen more than once.- Without repetition: n! ÷ (r! × (n − r)!), in practice the r numbers counting down from n divided by r!.
- With repetition the same formula is applied to n + r − 1 items (the “stars and bars” method).
- Everything uses exact integer arithmetic; results longer than 15 digits are summarised in scientific notation, with every digit in the table.
Example
A committee of 3 from 10 people: C(10, 3) = (10 × 9 × 8) ÷ 3! = 720 ÷ 6 = 120. There are C(49, 6) = 13,983,816 ways to pick 6 of 49 numbers, so one ticket wins with a chance of 0.00000715%. Three scoops from 5 flavours, repeats allowed: C(7, 3) = 35.
Frequently asked questions
What is the difference between combinations and permutations?
Order does not matter for combinations and does for permutations. A committee of 3 from 10 people: C(10, 3) = 120; a chair, deputy and secretary from the same 10: P(10, 3) = 720, six times as many because each committee can share the roles in 3! = 6 ways.
Why does C(n, r) equal C(n, n − r)?
Choosing r items is the same as choosing the n − r left behind. So C(10, 7) = C(10, 3) = 120, and the smaller one is the quicker sum.
When do I need combinations with repetition?
When the same kind can be picked more than once: 3 scoops from 5 flavours, 10 marbles from 4 colours. The formula is C(n + r − 1, r); for 3 scoops from 5 flavours, C(7, 3) = 35.
Why are C(0, 0) and C(n, 0) equal to 1?
There is exactly one way to choose nothing: the empty choice. If r > n (without repetition) no choice is possible and C(n, r) = 0; the tool flags it.