Combinations Calculator (nCr)

Calculate the number of ways to choose r items from a set of n items, where order doesn't matter.

Total Items (n)

Items to Choose (r)

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Combinations C(n,r)

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Permutations P(n,r)—

How it works

Combinations answer the question: 'In how many ways can I choose r items from a group of n, if the order I pick them doesn't matter?' This is written as C(n, r), nCr, or the binomial coefficient (n choose r). The key distinction is between combinations and permutations. If you're choosing 3 people from a group of 10 for a committee (where it doesn't matter who is first, second, or third), use combinations: C(10, 3) = 120. But if you're assigning 3 people to the roles of President, VP, and Treasurer (where order matters), use permutations: P(10, 3) = 720. Combinations appear everywhere in probability, statistics, and everyday life. Lottery odds are calculated using combinations (choosing 6 numbers from 49). In genetics, combinations count the number of possible gene combinations. In card games, combinations determine the number of possible hands. The formula uses factorials (n! = n × (n−1) × (n−2) × ... × 1). For large values, the numbers grow extremely quickly — C(50, 25) is over 126 trillion, which is why lotteries are so hard to win. Because order is ignored, a combinations count always comes out smaller than the matching permutation count, and that gap widens sharply as r grows.

Formula

C(n, r) = n! / (r! × (n − r)!)

The n! in the numerator counts every ordered arrangement of all items. Because a combination ignores order, you divide by r! to erase the different orderings of the r chosen items, and by (n−r)! to erase the orderings of the items left behind. What remains is the count of distinct groups. Permutations skip the r! step, since there order still matters, which is why they always exceed combinations.

Source:Concrete Mathematics, Graham, Knuth & Patashnik

We review all our tools from time to time to ensure accuracy.

Worked example

Total Items (n):10Items to Choose (r):3

Choosing 3 players from a squad of 10 for a mini-game: C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120 ways.

How to use this calculator

1
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2
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3
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Frequently asked questions

What is the difference between combinations and permutations?

The key difference is whether order matters. Combinations are used when the selection itself is what counts, regardless of the order items are arranged. Permutations are used when the order or arrangement of selected items matters. For example, choosing 3 flavors from 10 for a sampler plate is a combination (the plate is the same regardless of selection order). Arranging those 3 flavors in order from favorite to least favorite is a permutation. Permutations always equal or exceed combinations for the same n and r.

What does C(n, 0) and C(n, n) equal?

C(n, 0) always equals 1 — there is exactly one way to choose nothing from any set. Similarly, C(n, n) always equals 1 — there is exactly one way to choose everything. These are important edge cases in combinatorics. C(n, 1) = n, because there are n ways to choose a single item from n items. These base cases are fundamental to Pascal's Triangle, where each number is the sum of the two numbers above it.

How do combinations relate to Pascal's Triangle?

Pascal's Triangle is a direct representation of combinations. Each entry in the triangle at row n, position r equals C(n, r). For example, row 4 is: 1, 4, 6, 4, 1 — which are C(4,0)=1, C(4,1)=4, C(4,2)=6, C(4,3)=4, C(4,4)=1. Pascal's Triangle reveals the symmetry of combinations: C(n, r) = C(n, n−r). This symmetry makes sense intuitively: choosing r items to include is equivalent to choosing n−r items to exclude.

What is the binomial coefficient and how does it relate to combinations?

The binomial coefficient is just another name for C(n, r), written as (n choose r). It's called 'binomial' because it appears in the Binomial Theorem: (a + b)^n expands into a sum of terms, each with a binomial coefficient. For example, (a + b)^3 = C(3,0)a³ + C(3,1)a²b + C(3,2)ab² + C(3,3)b³ = a³ + 3a²b + 3ab² + b³. These coefficients (1, 3, 3, 1) match the third row of Pascal's Triangle.

How do I calculate combinations when n and r are very large?

Standard factorial calculations become impractical for large values because the numbers grow astronomically fast (100! has 158 digits). Practical approaches include: canceling common factors before multiplying (C(100, 3) = (100 × 99 × 98) / (3 × 2 × 1) rather than computing 100! fully), using logarithms of factorials for computation, or using specialized big-integer libraries. Many statistical software packages handle large combinations precisely. For approximate values, Stirling's approximation can estimate factorials: n! ≈ √(2πn)(n/e)^n.