Updated · Built and checked by the PlinzoTools team, formulas tested against published sources
Calculate nCr -- the number of unordered selections of r items from n.
10
A combination counts the number of ways to select r items from a set of n, where order doesn't matter (e.g. choosing 3 lottery numbers -- {1,2,3} is the same selection regardless of the order drawn).
Enter n (total items) and r (items chosen, r ≤ n). The number of combinations appears instantly.
nCr = n! / (r! × (n-r)!)
Choose combinations when you care only about which items are selected, not the order they were picked. Forming a three-person committee from a department, picking five cards from a deck, and selecting toppings for a pizza from a list are all combination problems. Two selections that contain the same items are the same combination, however they were arranged.
A hypothetical class of 12 students needs to send 4 representatives to a competition, with no special roles. Enter n = 12 and r = 4.
12C4 = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1) = 11,880 / 24 = 495
There are 495 different teams. If instead the four students needed distinct roles, such as captain and reserve, order would matter and you would use the permutation count, which is 11,880.
Rather than computing full factorials, the tool multiplies and divides step by step, and it automatically uses the smaller of r and n - r. That is valid because choosing r items to include is the same as choosing n - r items to leave out. For example, 50C48 has the same value as 50C2 = 1,225, which the tool can work out with just two steps. The arithmetic is done with exact integers, so there is no rounding at all.
Both n and r must be whole numbers, non-negative, with r no larger than n. Values like nC0 and nCn both equal 1, because there is exactly one way to choose nothing and exactly one way to choose everything. The tool accepts n up to 10,000 and always shows exact digits: results up to 40 digits are written in full with thousands separators, and longer ones appear in scientific notation with the exact leading digits (for example 10,000C5,000 has about 3,009 digits).
Combinations are the basis for many probability questions. If a hypothetical raffle draws 3 winners from 20 tickets, the number of possible winning groups is 20C3 = 1,140. If you hold one ticket, the chance of being in the winning group equals the number of groups containing you divided by all groups: 19C2 / 20C3 = 171 / 1,140 = 0.15, or 15 percent. Real games of chance have rules of their own, so check them before applying this model.
Because nCr equals nC(n-r), you can double-check your result by entering the complementary r. Both should match.
Combinations are always ≤ the equivalent permutation count, since they don't count reorderings of the same selection separately. See the Permutation Calculator.
Lottery odds, poker hand counting, and choosing committees/teams from a larger group are all classic combination problems.
Choosing which r items to take automatically decides which n - r items to leave out, so the two counts are always equal. This is a handy check on your answer.
That is 52C5 = 2,598,960 hands, since the order in which cards are dealt does not matter.
It equals 1. There is exactly one way to select nothing from a group.
No. It counts selections without repetition. Selections with repeats use a different formula.
Reference: NIST Weights and Measures