Binomial Coefficient Calculator
Compute C(n, k) — the number of ways to choose k items from n when order does not matter. Enter whole numbers for n and k and get the binomial coefficient, the order-matters permutation count P(n, k) for contrast, and the factorial breakdown.
Example: with n (total items) 10 · k (items chosen) 3 → Binomial coefficient C(n, k): 120.
- Permutations P(n, k) — order matters720
- SymmetryC(10, 3) = C(10, 7) = 120
- Computed as10! / (3! × 7!) = 120
Computed by the calculator below using its default values. Change any input to see your own numbers.
C(n, k) = n! / (k! (n − k)!) counts unordered selections. C(10, 3) = 120; ordering each selection 3! ways gives P(10, 3) = 720.
What n choose k actually counts
C(n, k) answers one question: from n distinct items, how many different groups of k can you form when the order inside the group is irrelevant? Choosing 3 pizza toppings from 10 gives C(10, 3) = 120 possible combinations. The formula n!/(k!(n − k)!) is a correction story: n!/(n − k)! counts ordered selections (720 for our toppings), and dividing by k! collapses the 3! = 6 orderings of each group into one.
The same numbers fill Pascal's triangle — row n, position k — and name the coefficients in (x + y)ⁿ, which is where 'binomial' comes from. Two identities fall straight out of the counting story: C(n, k) = C(n, n − k), because choosing 3 toppings to include is the same act as choosing 7 to leave out, and C(n, 0) = 1, because there is exactly one way to pick nothing.
How it’s calculated
C(n, k) = n! / (k!(n − k)!), evaluated with the multiplicative form C(n, k) = ∏ᵢ₌₁..ₖ (n − k + i)/i using k' = min(k, n − k) — this avoids computing giant factorials and stays exact for all results below 2⁵³. P(n, k) = n × (n − 1) × ... × (n − k + 1). Results at or above 10¹⁵ display in scientific notation; C(n, k) = 0 when k > n.
Assumes n distinct items chosen without repetition — combinations with repetition allowed use a different formula, C(n + k − 1, k).
Binomial coefficients you have already met
| Scenario | Coefficient | Value |
|---|---|---|
| 5-card poker hands from a 52-card deck | C(52, 5) | 2,598,960 |
| 6-number lottery picks from 49 | C(49, 6) | 13,983,816 |
| 3 pizza toppings from a menu of 10 | C(10, 3) | 120 |
| Ways 10 coin flips land exactly 5 heads | C(10, 5) | 252 |
Computed with C(n, k) = n!/(k!(n − k)!); the poker and lottery values are the standard published counts.
Common mistakes
- Using C(n, k) when order matters — awarding gold, silver, and bronze to 3 of 10 runners is P(10, 3) = 720, not 120.
- Computing the full factorials first: 52! has 68 digits and overflows most calculators, while the multiplicative form never leaves ordinary numbers.
- Treating C(n, 0) as 0 — there is exactly one way to choose nothing, so it equals 1.
- Reading k > n as an error — it is a valid question with answer 0, and the convention keeps binomial-theorem sums clean.
Frequently asked questions
What is the binomial coefficient formula?
C(n, k) = n! / (k!(n − k)!). For C(10, 3): 10!/(3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 120 — cancel the shared 7! before multiplying anything big.
What is the difference between combinations and permutations?
Combinations ignore order, permutations count it. P(n, k) = C(n, k) × k!, so P(10, 3) = 120 × 6 = 720. Committees are combinations; podium finishes are permutations.
How do you say and use C(n, k)?
Read it as 'n choose k'. It is the count of k-item subsets of n things, the entry in row n, position k of Pascal's triangle, and the coefficient of x^k y^(n−k) in the expansion of (x + y)ⁿ.
Why does C(n, k) equal C(n, n − k)?
Choosing which k items to take is the same decision as choosing which n − k to leave behind, so the counts must match: C(52, 5) = C(52, 47) = 2,598,960. It also means the fastest computation always uses the smaller of k and n − k.
Why is 0! equal to 1?
By convention, and the convention is forced: C(n, 0) must be 1 (one way to choose nothing), and the formula n!/(0! × n!) only gives 1 if 0! = 1. The empty product being 1 keeps all of combinatorics consistent.