n Choose k Calculator – Binomial Coefficient
Calculate the binomial coefficient C(n, k) — the number of ways to choose k items from n without regard to order. Enter n and k and get an instant result with the formula and a worked example.
Calculate “n Choose k” Combinations
Binomial Coefficient Formula
Where:
- \(n\) — total number of items (a non-negative integer)
- \(k\) — number of items to choose (\(0 \le k \le n\))
- \(n!\) — factorial of \(n\), the product \(1 \times 2 \times \dots \times n\), with \(0! = 1\)
The formula counts how many distinct subsets of size \(k\) can be formed from \(n\) items when order does not matter. It is the same coefficient that appears in Pascal's Triangle and in the expansion of \((a+b)^n\).
Binomial Coefficient – Calculation Example
Example: a committee of 2 people must be chosen from a group of 5. How many different committees are possible? Here \(n = 5\) and \(k = 2\).
\[ \binom{5}{2} = \frac{5!}{2!\,(5-2)!} = \frac{120}{2 \times 6} = \frac{120}{12} = 10 \]
Result: \(\binom{5}{2} = 10\) — there are 10 ways to choose 2 people out of 5.
How to Use the Calculator
- Enter the total number of items in the field n (total items).
- Enter the number of items to choose in the field k (selected items). The value of \(k\) cannot exceed \(n\).
- Read the result under Binomial Coefficient — this is \(\binom{n}{k}\), the number of possible selections.
What Is the Binomial Coefficient?
The binomial coefficient \(\binom{n}{k}\), read aloud as "n choose k", answers a fundamental counting question: in how many ways can you pick \(k\) objects from a set of \(n\) objects when the order of the picked objects does not matter? If you have 5 books and want to take 2 on a trip, it does not matter which one you grab first — what matters is which pair ends up in your bag. That is why binomial coefficients count combinations, as opposed to permutations, where order matters and the count would be larger.
The name comes from the binomial theorem: the coefficients in the expansion of \((a+b)^n\) are exactly \(\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}\). For example, \((a+b)^2 = a^2 + 2ab + b^2\), where the coefficient 2 equals \(\binom{2}{1}\). If you want to expand products of two-term expressions, the Multiplying Binomials calculator and the FOIL method calculator handle that step directly.
Binomial coefficients are also the building blocks of probability. If an experiment has \(n\) independent trials, each succeeding with probability \(p\), the chance of exactly \(k\) successes is \(\binom{n}{k} p^k (1-p)^{n-k}\). The coefficient counts which of the \(n\) trials are the successful ones, which is why it appears whenever outcomes are counted in groups.
Each row of Pascal's Triangle lists the coefficients for one value of \(n\): row 5 is 1, 5, 10, 10, 5, 1, which contains \(\binom{5}{2} = 10\). The triangle makes the symmetry \(\binom{n}{k} = \binom{n}{n-k}\) and Pascal's rule, \(\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\), easy to see. The same triangle also connects to the Fibonacci series, since the sums of its shallow diagonals produce the Fibonacci numbers.
Interpreting the Result
The result is always a whole number, because you cannot have a fraction of a selection. Two special cases are worth remembering: \(\binom{n}{0} = 1\) (there is exactly one way to choose nothing) and \(\binom{n}{n} = 1\) (one way to choose everything). If \(k > n\), the selection is impossible and the calculator shows an error rather than a number.
Useful identities to check your work: \(\binom{n}{1} = n\), \(\binom{n}{2} = \frac{n(n-1)}{2}\), and by symmetry \(\binom{n}{k} = \binom{n}{n-k}\), so \(\binom{5}{3} = \binom{5}{2} = 10\).
Assumptions and Limitations
- Integers only. Both \(n\) and \(k\) must be non-negative integers. The combinatorial meaning of "choosing items" does not extend to fractional inputs.
- \(k \le n\). If the number of items to choose exceeds the number available, the calculator reports an error; mathematically, \(\binom{n}{k} = 0\) for \(k > n\).
- Order is ignored. This calculator counts unordered selections. If the order of the chosen items matters, you need permutations, \(P(n,k) = \frac{n!}{(n-k)!}\), which is always greater than or equal to \(\binom{n}{k}\).
- No repetition. Each of the \(n\) items can be chosen at most once. Selections with repetition allowed use a different formula, \(\binom{n+k-1}{k}\).
- Size limits. Factorials grow extremely fast; \(70!\) already exceeds \(10^{100}\). For very large \(n\), results may be limited by number precision even though the mathematical value is exact.
Common Values Reference
| \(n\) | \(k\) | \(\binom{n}{k}\) | Interpretation |
|---|---|---|---|
| 4 | 2 | 6 | Pairs from 4 items |
| 5 | 2 | 10 | Pairs from 5 items |
| 6 | 3 | 20 | Triples from 6 items |
| 10 | 5 | 252 | Half of 10 items |
| 52 | 5 | 2,598,960 | 5-card poker hands |
The last row shows why binomial coefficients are the standard tool for card problems: the number of distinct 5-card hands from a 52-card deck is exactly \(\binom{52}{5}\).
Frequently Asked Questions
How do I calculate a binomial coefficient by hand without computing huge factorials?
Use the multiplicative shortcut: \(\binom{n}{k} = \frac{n}{1} \times \frac{n-1}{2} \times \frac{n-2}{3} \times \dots\), with exactly \(k\) factors on top and bottom. For \(\binom{8}{3}\): \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\). Multiply and divide as you go to keep the intermediate numbers small. Also exploit symmetry: compute with the smaller of \(k\) and \(n-k\).
What notations mean the same thing as C(n, k)?
All of these refer to the same quantity: \(C(n,k)\), \({}_nC_k\), \(C^n_k\), \(C^k_n\), and \(\binom{n}{k}\). They are read "n choose k" or "n taken k at a time." The notation \(\binom{n}{k}\) is standard in higher mathematics, while \({}_nC_k\) and \(C(n,k)\) are common in school textbooks and on scientific calculators, which often have a dedicated nCr button.
Why is C(n, k) smaller than the number of permutations P(n, k)?
Permutations count ordered arrangements, while binomial coefficients count unordered selections. Each unordered group of \(k\) items corresponds to \(k!\) different orderings, so \(P(n,k) = \binom{n}{k} \times k!\). For example, picking 2 winners from 8 contestants in order gives \(8 \times 7 = 56\) outcomes, but if the order does not matter there are only \(\binom{8}{2} = 28\) pairs.
What happens if k = 0 or k = n?
Both cases give exactly 1. There is only one way to choose nothing (the empty selection) and only one way to take everything. These are the first and last entries of every row in Pascal's Triangle, which always start and end with 1. So \(\binom{7}{0} = \binom{7}{7} = 1\), no matter how large \(n\) is.
Can C(n, k) be negative or a fraction?
No. When \(n\) and \(k\) are non-negative integers with \(k \le n\), the result is always a non-negative whole number, because it counts something. If \(k > n\) the convention is \(\binom{n}{k} = 0\) — there are zero ways to pick more items than you have. Negative or fractional inputs do have values in advanced mathematics (generalized binomial coefficients), but they are not selection counts.
How is the binomial coefficient used in probability?
It appears in the binomial distribution: the probability of exactly \(k\) successes in \(n\) independent trials, each with success probability \(p\), is \(\binom{n}{k} p^k (1-p)^{n-k}\). The coefficient counts which trials are successful. For instance, the chance of exactly 3 heads in 10 fair coin flips is \(\binom{10}{3} (0.5)^{10} \approx 0.117\).
How large can n be before the result becomes unreliable?
Mathematically the value is always an exact integer, but factorials grow astronomically — \(70!\) already exceeds \(10^{100}\), and \(\binom{100}{50}\) is a 30-digit number. Standard floating-point arithmetic loses precision or overflows well before that. If you need exact values for large \(n\), use software with arbitrary-precision integers (Python, WolframAlpha) or the multiplicative method with exact integer division.