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A combination counts selections where "order doesn't matter": choosing r items from n, it makes no difference which order you pick them — {A, B} and {B, A} are the same selection. From a lottery ticket to forming a committee, from picking pizza toppings to card-hand probability, the answer to "how many different groups" is a combination.
Enter the number of items (n) and how many to choose (r), and C(n, r) is calculated instantly.
How is it calculated?
Formula
C(n, r) = n! ÷ [r! × (n−r)!]. Because order doesn't matter, we divide the permutation by r! (counting each group's internal orderings only once).
Combination or permutation?
The single question: does order matter? - If it does NOT → combination (team selection, lottery, a bouquet). - If it does → permutation (race positions, a password, president-and-deputy).
That is why C(n,r) ≤ P(n,r) always; the combination is smaller because it collapses orderings into one.
Useful properties
- C(n, 0) = C(n, n) = 1 (there is one way to choose nothing or everything).
- C(n, r) = C(n, n−r) (choosing r is the same count as choosing the n−r you leave behind).
Where it is used
- Games of chance: how many ways to pick 6 numbers from 49 (lottery odds).
- Grouping: selecting a committee, team or jury from a class.
- Probability: card-hand combinations, quality-control samples.
For ordered selections use a permutation calculator; for the factorial underneath use a factorial calculator.
Worked example
A committee of 3 is to be chosen from 10 students: since order doesn't matter (who was picked first is irrelevant), this is a combination. C(10, 3) = 10! ÷ (3! × 7!) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120 different committees. If instead they were chosen as chair-deputy-secretary (ordered), the answer would be P(10,3) = 720 — which shows why the combination is exactly one-sixth (3! = 6): each committee has 6 internal orderings, and the combination counts them once. In the lottery case, C(49, 6) = 13,983,816 different tickets.
FAQ
How is a combination calculated?+
With C(n, r) = n! ÷ [r! × (n−r)!] — the number of ways to choose r items from n without regard to order. Just enter n and r.
What is the difference between a combination and a permutation?+
In a combination order doesn't matter ({A,B}={B,A}); in a permutation it does. So C(n,r) = P(n,r) ÷ r!, and a combination is always less than or equal to a permutation.
Are lottery odds found with combinations?+
Yes — choosing 6 numbers from 49 is order-independent, so it is a combination: C(49,6) = 13,983,816. The odds of the jackpot are the reciprocal (1 in 13,983,816).
Why is C(n, 0) equal to 1?+
There is exactly one way to choose no items (the empty set). Similarly C(n, n) = 1: there is one way to choose all of them.
What does C(n, r) = C(n, n−r) mean?+
Choosing r items is the same count as choosing the n−r you leave out. For example C(10,3) = C(10,7) = 120 — a symmetry that simplifies calculation.
Which tool do I use if order matters?+
If order matters (race finish, password, titled roles), you need a permutation. Use a permutation calculator to find P(n, r).