Binomial Coefficient Calculator
The binomial coefficient C(n, k), read “n choose k,” counts how many ways you can pick k items from a set of n when the order of the pick does not matter. It is the number behind a lot of everyday counting: there are C(52, 5) = 2,598,960 possible 5-card hands in a standard deck, and C(49, 6) = 13,983,816 ways to choose six lottery numbers from 49. The name comes from algebra: C(n, k) is also the coefficient of the x^k term when (1 + x)^n is expanded.
The same number travels under several names, and they all land here. Textbooks that write the second input as r call it “n choose r”; this page writes it as k and calls it “n choose k.” Calculator buttons shorten both to nCr, and algebra texts call the result the binomial coefficient. The letter in the second slot changes nothing: 10 choose 3, C(10, 3), and 10C3 are one question with one answer, 120. Throughout this page, k plays the role that r plays in “n choose r” questions, so a choose calculator search and a binomial coefficient search end at the same tool.
The Binomial Coefficient Calculator below computes that number exactly, using a divide-at-every-step method that stays precise long after the naive “work out three factorials and divide” approach has overflowed. It also covers the two counting questions people confuse with it: permutations P(n, k), for when order does matter, and combinations with repetition C(n + k – 1, k), for when the same item can be picked more than once. Every result comes with the working shown, the full Pascal’s triangle row for your n, and a plain-English sentence stating what the number counts.
Results are exact integers up to about 9 x 10^15; beyond that they are shown in exponential form. n is capped at 1000 to keep the computation instant.
Quick Answer
Key takeaways
- C(n, k) = n! / (k! x (n - k)!). Order is the only difference between a combination and a permutation.
- C(n, k) always equals C(n, n - k): choosing who is in is the same count as choosing who is out.
- The calculator divides at every step of the multiplication, so every intermediate value stays an exact integer and factorials never need to be formed.
- Results are exact up to about 9 x 10^15. Above that the answer switches to exponential form, and the result badge tells you when that has happened.
- Quick Answer
- Combinations, permutations, or repetition: picking the right mode
- The formula, and why this calculator divides at every step
- Worked example: choosing 3 students from a class of 20
- Edge cases worth knowing before an exam or an audit
- Pascal's triangle, symmetry, and the row under every result
- Where the exactness ends, stated plainly
- What competing calculators miss
- Frequently asked questions
- Add This Calculator to Your Website
Combinations, permutations, or repetition: picking the right mode
Answer one question first: does reordering your pick create a different outcome? If a committee of Ana, Ben, and Cy is the same committee in any order, you want combinations. If first, second, and third place are different outcomes, you want permutations. If the same item can be picked repeatedly, like scoops from ice-cream flavours that nobody is removing from the tub, you want the repetition mode.
| Question | Mode | Formula | Example |
|---|---|---|---|
| Pick a group; order does not matter | Combinations C(n, k) | n! / (k! (n - k)!) | 5-card hand from 52 cards: 2,598,960 |
| Arrange or rank picks; order matters | Permutations P(n, k) | n! / (n - k)! | Gold, silver, bronze from 52 athletes: 52 x 51 x 50 = 132,600 |
| Pick with repeats allowed; order does not matter | Combinations with repetition | (n + k - 1)! / (k! (n - 1)!) | 3 scoops from 10 flavours: C(12, 3) = 220 |
Permutations always outnumber combinations for the same n and k (when both are above 1), because every chosen set of k items can be arranged in k! orders. That is exactly the relationship the formulas show: P(n, k) = C(n, k) x k!. If your permutation answer is not a multiple of k! relative to the combination answer, one of the two runs used the wrong mode. These counts are also the raw material of probability: a compound probability calculator divides favourable counts by total counts, and the total count is usually a binomial coefficient.
The formula, and why this calculator divides at every step
The definition is a ratio of factorials, and the calculator evaluates it in a way that never builds the factorials themselves. That choice matters more than it looks: 52! is a 68-digit number, and 100! has 158 digits, so a tool that literally computes n! first throws away exactness long before it divides.
Shrink k by symmetry. The calculator replaces k with min(k, n - k). C(52, 47) becomes C(52, 5): same answer, a tenth of the work.
Multiply and divide in lockstep. Starting from 1, for each i from 1 to the reduced k, it multiplies by the next descending term and divides by i immediately, rounding to clear floating point dust. After every division the running value is a genuine integer: it is itself a binomial coefficient.
Stop while the number is still exact. Because no factorial is ever formed, the computation stays exact until the final answer itself passes what a standard number can hold precisely, about 9 x 10^15. Past that point the display switches to exponential form and says so.
The steps list under each result mirrors this: it shows the symmetry reduction, the multiplicative form, and, for n up to 18, the full factorial breakdown (for example 10! = 3,628,800) so a student can verify the answer against the textbook method digit for digit. Above n = 18 the factorial values stop being readable, so the steps say that and rely on the multiplicative form instead.
Worked example: choosing 3 students from a class of 20
Enter n = 20 and k = 3 in combinations mode. Every figure below is what the calculator returns, and each one can be checked by hand in under a minute, which is the standard these examples should meet.
| Step | Working | Value |
|---|---|---|
| Factorial form | 20! / (3! x 17!) | the 17! cancels most of 20! |
| Multiplicative form | (20 x 19 x 18) / (3 x 2 x 1) | 6,840 / 6 |
| Result | C(20, 3) | 1,140 ways |
| Same inputs, permutations | 20 x 19 x 18 | P(20, 3) = 6,840 |
| Same inputs, with repetition | C(20 + 3 - 1, 3) = C(22, 3) | 1,540 |
| Symmetry check | C(20, 17) should equal C(20, 3) | 1,140 |
The repetition figure deserves a second look. C(22, 3) = (22 x 21 x 20) / 6 = 9,240 / 6 = 1,540. It is larger than 1,140, as it must be: allowing repeats adds outcomes (the same student picked twice is not a real committee, but in the flavour-scoop world it is a real order). If a repetition answer ever comes out smaller than the plain combination for the same n and k, the inputs were swapped.
A smaller classroom pair is worth keeping, because it is the fastest hand check of the mode choice. A committee of 3 from 10 students is 10 choose 3: C(10, 3) = (10 x 9 x 8) / (3 x 2 x 1) = 720 / 6 = 120. Give those same 3 students distinct jobs, chair, secretary, and treasurer, and order starts to matter: P(10, 3) = 10 x 9 x 8 = 720, exactly 3! = 6 times the committee count. One step smaller, choosing 3 students from 7 gives C(7, 3) = (7 x 6 x 5) / 6 = 35. Run any of the three above and the steps under the result show the same multiplicative form.
Edge cases worth knowing before an exam or an audit
The boundary inputs are where calculators quietly differ, so here is exactly how this one behaves. Each line was verified against the tool's code, not assumed from the formula.
| Input | Result | Why |
|---|---|---|
| k = 0, any n | C(n, 0) = 1 | There is exactly one way to choose nothing. |
| k = n | C(n, n) = 1 | Exactly one way to take everything. |
| n = 0, k = 0 | 1 | The empty choice from an empty set. |
| k = 1 | C(n, 1) = n | One choice per item in the set. |
| k = n - 1 | C(n, n - 1) = n | Choosing who to leave out: n candidates for the omission. |
| k greater than n (combinations or permutations) | Error, no result | You cannot choose more distinct items than exist. In repetition mode k may exceed n, because repeats are allowed. |
| Very large n, e.g. C(1000, 500) | About 2.70 x 10^299, exponential form | The exact integer has 300 digits; beyond about 9 x 10^15 the tool reports magnitude instead of pretending to digit-perfect precision. |
| Negative or blank n or k | Error, no result | Counts of items cannot be negative, and blanks are rejected rather than treated as zero. |
One behaviour to be aware of: decimal entries are truncated to whole numbers by the input parser (5.9 is read as 5). Combinations only make sense for whole items, so enter integers and the truncation never comes into play. If you are feeding these counts into a significance test afterwards, the statistical significance calculator is the natural next step for proportion-based tests.
Pascal's triangle, symmetry, and the row under every result
Every result includes row n of Pascal's triangle, the row whose entries are C(n, 0), C(n, 1), through C(n, n). For n = 10 the row is 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. Read it and three properties become visible without any algebra. The row is a mirror image, which is the symmetry C(n, k) = C(n, n - k). The entries sum to 2^n, here 1,024, because every subset of the 10 items is counted exactly once across the row. And each entry is the sum of the two above it, Pascal's rule, which is how the calculator's row builder works. Rows are shown for n up to 60; beyond that the row would be wider than the page and larger than exact display allows.
The same counting sizes search spaces in computing. A password or a combination lock with a space of C(n, k) outcomes has a very different strength from one with P(n, k), and the brute force calculator uses counts of exactly this kind to estimate how long exhaustive guessing would take.
Where the exactness ends, stated plainly
This tool computes with ordinary double-precision numbers, not arbitrary-precision integers. That is a real limit and it is printed here rather than buried: the divide-at-every-step method keeps every intermediate an integer, but once the final answer exceeds about 9 x 10^15 (2^53), a double cannot represent every integer, and the display moves to exponential form with six decimal digits, for example 2.702882 x 10^299 style output for the largest inputs. The verdict badge under the result tells you which regime you are in. For classroom work, card and lottery counting, committee problems, and probability denominators, answers sit far below the ceiling and are digit-exact.
What competing calculators miss
The strongest rivals each do one thing well. Pearson's combination calculator uses arbitrary-precision arithmetic and has no exactness ceiling at all; if you need all 300 digits of C(1000, 500), use that class of tool. Calculator.now offers a choice of computation methods, which is genuinely instructive. OnlyCalculators matches this tool mode for mode. What the set generally lacks is the combination of all three modes, the working, the Pascal row, and an honest statement of limits on one page.
Winning move 1: three modes, one input pair. Most tools split combinations, permutations, and repetition across separate pages, so comparing them means re-entering numbers and trusting three different implementations. Here the same n and k flow through all three formulas, and the with-repetition value appears under every result for instant comparison.
Winning move 2: the working is the default, not an option. Symmetry reduction, multiplicative form, and a factorial breakdown for classroom-sized n are shown on every run. One ranking rival's worked example even contradicts itself, computing C(10, 3) = 120 in its formula and then calling the same problem 210 in its text. Steps that are actually consistent with the answer are the whole point of showing steps.
Winning move 3: limits in the open. The exact-integer ceiling, the n cap of 1,000, the Pascal row limit of 60, and the factorial breakdown limit of 18 are all stated on the page and enforced with plain error messages, including the case rivals often fumble: k greater than n is rejected for combinations and permutations, and allowed only where it makes sense, in repetition mode.
Frequently asked questions
C(n, k) counts groups where order does not matter; P(n, k) counts arrangements where it does. For n = 20, k = 3: C(20, 3) = 1,140 groups, P(20, 3) = 6,840 arrangements. The ratio between them is always k!, here 6, because each group of 3 can be arranged 6 ways.
nCr is calculator shorthand for C(n, r): the number of ways to choose r items from n without regard to order. It is the same number as the binomial coefficient and as "n choose r." The corresponding shorthand for permutations is nPr. This calculator labels the modes in words so the shorthand is never the only explanation.
It is the number of ways to choose r items from a set of n when the order of choosing does not matter, computed as n! / (r! x (n - r)!). On this page the same count is written C(n, k): r and k are two names for the number chosen. Choosing a committee of 3 from 10 students is 10 choose 3, which equals 120, and entering n = 10 and k = 3 in combinations mode above returns exactly that.
Because (52 x 51 x 50 x 49 x 48) / (5 x 4 x 3 x 2 x 1) = 311,875,200 / 120 = 2,598,960. The numerator counts ordered deals of 5 cards; dividing by 5! = 120 removes the 120 orders in which the same hand can arrive, leaving the number of distinct hands.
For combinations and permutations, no: you cannot pick more distinct items than the set contains, and the calculator returns an error. For combinations with repetition, yes: repeats let you pick 8 items from 5 types, giving C(5 + 8 - 1, 8) = C(12, 8) = 495.
The count rises, because picks like (A, A, B) become possible. The formula becomes C(n + k - 1, k), the stars-and-bars result: distributing k identical picks among n types. Three scoops from 10 flavours give C(12, 3) = 220, against C(10, 3) = 120 when each flavour can be used once.
Because the exact answer passed about 9 x 10^15, the largest range in which ordinary computer numbers represent every integer exactly. The tool reports the magnitude in exponential form and the result badge changes to say so, rather than printing a long digit string whose last digits would be wrong.
Yes. Choosing k items to include is the same act as choosing n - k items to leave out, so the counts must match: C(20, 3) = C(20, 17) = 1,140. The calculator exploits this to do less work, always computing with the smaller of k and n - k.
It counts equally likely outcomes, which is the denominator of most exact probability fractions. A 6/49 lottery has C(49, 6) = 13,983,816 combinations, so the chance of one specific ticket matching all six numbers is 1 in 13,983,816. Binomial probabilities for repeated trials are built from the same coefficient, multiplied by powers of the success and failure probabilities.
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References
- Weisstein, Eric W., "Binomial Coefficient," MathWorld, Wolfram Research: the factorial definition, symmetry, and Pascal's rule used by this calculator. MathWorld: Binomial Coefficient. Accessed October 2026.
- National Institute of Standards and Technology, Digital Library of Mathematical Functions, Chapter 26 "Combinatorics," section 26.3: the formal definition of the binomial coefficient and its identities. NIST DLMF: Binomial Coefficients (26.3). Accessed October 2026.
- National Institute of Standards and Technology, NIST/SEMATECH e-Handbook of Statistical Methods, "Binomial Distribution": how binomial coefficients enter the binomial probability mass function. NIST e-Handbook: Binomial Distribution. Accessed October 2026.
- OpenStax, Introductory Statistics 2e, section 3.5 "Tree and Venn Diagrams," which builds counting outcomes toward probability denominators. OpenStax: Tree and Venn Diagrams. Accessed October 2026.
- Khan Academy, "Factorial and counting seat arrangements," the permutations lesson whose counting argument distinguishes ordered arrangements from combinations. Khan Academy: Factorial and Counting Seat Arrangements. Accessed October 2026.