Binomial Coefficient Calculator (nCr)
Enter n (total items) and k (items chosen) to calculate the binomial coefficient C(n,k), also written nCr. You get the combination count, permutation count, the with-repetition variant, a full factorial step-by-step breakdown, and the row of Pascal's triangle that contains your result. Results update as you type.
Formula
Worked example
How many 3-person committees can be chosen from a group of 10? n=10, k=3. C(10,3) = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120 committees. If the roles were ranked (chair, vice-chair, secretary), order matters and the answer would be P(10,3) = 10 x 9 x 8 = 720.
What is the binomial coefficient C(n,k)?
The binomial coefficient C(n,k), read "n choose k" and written nCr on calculators, counts the number of ways to select k items from a set of n distinct items when the order of selection does not matter. The formula is C(n,k) = n! / (k! x (n-k)!), where n! (n factorial) means the product of every integer from 1 to n. The name "binomial" comes from the binomial theorem, which states that (x+y)^n expands into a sum of terms whose coefficients are exactly the values C(n,0), C(n,1), ..., C(n,n). These same numbers form the rows of Pascal's triangle: row n lists C(n,0) through C(n,n), and each entry is the sum of the two directly above it.
Combinations vs. permutations - when does order matter?
The key distinction is whether the arrangement matters. If you are picking a 5-card poker hand, the order the cards arrive in is irrelevant and you use C(52,5) = 2,598,960. If you are awarding a gold, silver, and bronze medal to 3 of 10 runners, the order absolutely matters and you use P(10,3) = 720. The relationship is P(n,k) = C(n,k) x k!, because for every combination of k items there are k! ways to arrange them. A useful mnemonic: Combinations are for Committees (unordered groups); Permutations are for Podiums (ordered ranks).
Combinations with repetition (stars and bars)
When repetition is allowed - for example, choosing 3 scoops from 5 ice-cream flavours where the same flavour can be picked more than once - the count is C(n+k-1, k), sometimes called "multichoose". This formula comes from the stars-and-bars argument: imagine k indistinguishable stars placed into n bins separated by n-1 bars; you are choosing where to put the k stars among k+(n-1) positions. For 5 flavours and 3 scoops: C(5+3-1, 3) = C(7,3) = 35.
Pascal's triangle and binomial expansion
Pascal's triangle is formed by writing 1 at the apex and making each subsequent entry the sum of the two entries above it. Row n (starting at row 0) contains C(n,0), C(n,1), ..., C(n,n). The triangle captures several identities at once: Pascal's identity C(n,k) = C(n-1,k-1) + C(n-1,k); the symmetry C(n,k) = C(n,n-k); and the row-sum identity C(n,0)+C(n,1)+...+C(n,n) = 2^n (the total number of subsets of an n-element set). In the binomial expansion (x+y)^n = sum of C(n,k) x^(n-k) y^k for k=0..n, each coefficient is read directly from row n of Pascal's triangle.
Pascal's triangle - rows 0 through 8
| n (row) | C(n,0) | C(n,1) | C(n,2) | C(n,3) | C(n,4) | C(n,5) | C(n,6) | C(n,7) | C(n,8) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | ||||||||
| 1 | 1 | 1 | |||||||
| 2 | 1 | 2 | 1 | ||||||
| 3 | 1 | 3 | 3 | 1 | |||||
| 4 | 1 | 4 | 6 | 4 | 1 | ||||
| 5 | 1 | 5 | 10 | 10 | 5 | 1 | |||
| 6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||
| 7 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | |
| 8 | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 |
Each entry is C(row, position). Row n contains all binomial coefficients for that n, and each entry equals the sum of the two entries directly above it.
Frequently asked questions
What does C(n,k) mean and how do you read it?
C(n,k) is read "n choose k". It is the number of ways to select k items from a collection of n distinct items without regard to order. On a calculator it appears as the nCr button. The formula is n! divided by (k! times (n-k)!). For example, C(6,2) = 6!/(2! x 4!) = 720/(2 x 24) = 15, meaning there are 15 ways to pick any 2 items from a set of 6.
What is the difference between C(n,k) and P(n,k)?
C(n,k) counts selections where the order does not matter (committees, card hands, lottery numbers). P(n,k) counts arrangements where the order does matter (race podiums, PIN codes with distinct digits, ranked lists). P(n,k) = C(n,k) times k!, so for the same n and k, the permutation count is always larger than or equal to the combination count.
Why is C(n,k) = C(n, n-k)?
Choosing which k items to include is equivalent to choosing which (n-k) items to leave out - both decisions fully determine the same subset. Formally, swapping k and (n-k) in the formula n!/(k!(n-k)!) yields the identical expression. This symmetry means C(10,3) = C(10,7) = 120, and it is visible in Pascal's triangle where each row reads the same forwards and backwards.
What is Pascal's triangle and how does it relate to C(n,k)?
Pascal's triangle is an infinite triangular array where row n (counting from 0) lists exactly the values C(n,0), C(n,1), ..., C(n,n). Each interior entry is the sum of the two entries directly above it, which reflects Pascal's identity: C(n,k) = C(n-1,k-1) + C(n-1,k). The triangle is useful for quick lookups: row 5 is 1, 5, 10, 10, 5, 1, so C(5,2) = 10 by reading the third entry. The row sums equal powers of 2: row n sums to 2^n.
How large can n be in this calculator?
This tool handles n up to 1000. For n up to about 20 it displays full factorial expansions; for larger n it uses a multiplicative shortcut that avoids computing enormous factorials directly. Results are rounded to the nearest integer. For very large n (above a few hundred) with k near n/2, the values become astronomically large (hundreds of digits) and are returned as approximate integers.
When should I use combinations with repetition?
Use C(n+k-1, k) when the same item can be chosen more than once and order still does not matter. Classic examples: choosing 3 toppings from a menu of 5 where duplicates are fine (extra cheese counts twice); distributing 10 identical balls into 4 boxes; rolling two identical dice and recording only which numbers appeared. When each item can be used at most once, revert to the standard C(n,k) formula.