Combination Calculator

Combinations with repetition

Sometimes you choose items where the same type can be picked more than once, such as choosing 3 scoops from 5 ice-cream flavours where two scoops can share a flavour. The formula is C(n + r - 1, r). Here that is C(7, 3) = 35, compared with C(5, 3) = 10 when every scoop must be a different flavour.

The idea behind it, often called stars and bars, is to treat the r picks as stars and use n - 1 bars to split them into flavour groups, which gives r + n - 1 positions in which to place the r stars.

Combinations: choosing without order

A combination counts the ways to choose r items from n when the order does not matter. The formula is C(n, r) = n! ÷ (r! × (n − r)!), where n! (n factorial) is n × (n − 1) × … × 1.

Choosing 3 committee members from 10 people gives C(10, 3) = 10! ÷ (3! × 7!) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120 possible committees. In practice you cancel the large factorials instead of computing them in full.

Combinations versus permutations

If order matters — president, secretary and treasurer rather than three equal members — you count permutations instead: P(10, 3) = 720. Each group of 3 can be arranged in 3! = 6 orders, which is exactly why C = P ÷ r!.

Ask yourself: would swapping two chosen items give a different outcome? If not, use combinations.

Real examples

A 6-from-49 lottery has C(49, 6) = 13,983,816 possible tickets, so one ticket has about a 1 in 14 million chance of the jackpot. A pizza shop with 8 toppings offers C(8, 2) = 28 different two-topping pizzas. Combinations also give the coefficients in the binomial expansion and the probabilities in the binomial distribution. A useful symmetry: C(n, r) = C(n, n − r), so choosing 3 people to include is the same count as choosing 7 to leave out.