Binomial Coefficient Calculator

Why the factorial formula fails, twice

n! / (k!(n−k)!) is the definition and a poor algorithm, for two independent reasons.

Written as 200!/(100!·100!), this cannot be computed in floating point at all: 200! is beyond the largest double (the ceiling is 170!), so the numerator and denominator are both Infinity and the ratio is NaN. The multiplicative form above never builds a number larger than the answer itself, which is 59 digits — well within reach.

The second reason applies long before the first. Even when the factorials fit, they are enormously larger than the answer: for C(52, 5) the answer is 2,598,960 while 52! has 68 digits. Building a 68-digit number to produce a 7-digit one is work you did not need to do, and in fixed-precision arithmetic it is where the rounding comes from.

49! is 63 digits long, while the answer is only 8. Computing the factorials first means handling numbers vastly larger than the result, for no benefit — the multiplicative form avoids them entirely.

The multiplicative form, and why it stays whole

Compute C(n, k) as (n−k+1)/1 × (n−k+2)/2 × … × n/k, alternating a multiplication and a division. The running total never exceeds the answer, and every division comes out exactly.

That exactness is not luck. After i steps the running total is C(n−k+i, i), which is an integer — and the underlying fact is that the product of any i consecutive integers is divisible by i!. So the division is always whole, and no rounding is ever needed at any point.

The tool shows those steps for 5 choose 2 and larger cases: 4 → 10. Each intermediate is a real binomial coefficient in its own right.

The symmetry that halves the work

C(1000, 999) = C(1000, 1), so this was computed with k = 1 instead of 999 — 1 multiplications rather than 999. Choosing 999 to keep is the same act as choosing 1 to leave out.

The reason is a bijection rather than an algebraic identity: choosing which k to take is the same act as choosing which n−k to leave. Every selection corresponds to exactly one rejection, so the two counts are equal.

Computationally it matters a great deal. C(1000, 999) computed directly would need 999 multiplications; using the symmetry it needs 1. The tool always uses the smaller index.

It is also why Pascal’s triangle is symmetric, and why the largest entry in a row is in the middle: k is exactly half of n, which is where C(n, k) is at its largest — the middle of the row is always the peak.

Pascal’s rule, and the triangle it builds

C(n, k) = C(n−1, k−1) + C(n−1, k). For 52 choose 5 that is 249,900 + 2,349,060 = 2,598,960.

The proof is one line of counting. Pick one particular object. Either it is in your selection — and then you are choosing the remaining k−1 from the other n−1 — or it is not, and you are choosing all k from the other n−1. Those cases do not overlap and cover everything.

That recurrence IS Pascal’s triangle: each entry is the sum of the two above it. And the rows sum to powers of two — The row sums to 2^5 = 32, because choosing SOME subset of 5 things means choosing 0 or 1 or … or 5 of them, and there are 2^5 subsets altogether.

Combinations or permutations, and how to tell

P(10, 5) = 30,240 against C(10, 5) = 252. The ratio is exactly 5! = 120, which is the number of orders the same 5 chosen items can appear in. Permutations count those separately; combinations do not. If the question says "arrange", "order", "rank" or "podium", it is a permutation; if it says "select", "team", "committee" or "hand", it is a combination.

The factor between them is exactly k!, which is the number of orders the same k chosen objects can appear in. Permutations count those separately; combinations do not. Nothing else differs.

So P(n, k) = k! × C(n, k), and if you have computed one you have effectively computed the other. For 52 choose 5 the combination count is 2,598,960 and the permutation count is 311,875,200 — a factor of 120 apart.

The wording test is reliable: a poker HAND is a combination, because the order the cards were dealt does not change the hand. A podium finish is a permutation, because first and second are different outcomes.

Three things the same number counts

C(52, 5) = 2,598,960 is the number of ways to choose 5 items out of 52 when the order of the choice does not matter. It is also the coefficient of x^5y^47 in the expansion of (x + y)^52, and the number of paths from the top of Pascal's triangle to this position — three descriptions of one number, and the reason it shows up in so many different questions.

Those three descriptions — a count of selections, a coefficient in a binomial expansion, and a count of lattice paths — are not analogies. They are the same number, and a fact proved in any one of the three settings holds in the other two.

Pascal’s rule is easiest to see as paths: a route to a given point either arrives from the left or from above. The row sum 2ⁿ is easiest as selections: choosing SOME subset of n objects means choosing 0 or 1 or … or n of them, and there are 2ⁿ subsets altogether — which is exactly what the Power Set Calculator enumerates.

C(6, 0) = 1: there is exactly one way to choose nothing, and that is why the ends of every row of Pascal's triangle are 1.

Sources and methodology

Binomial coefficients and Pascal’s rule are standard; the reference below carries the canonical statements.

• NIST Digital Library of Mathematical Functions — §26.3, Binomial Coefficients
• How we calculate — methodology

Method. Every figure on this page comes from src/lib/binomial-coefficient.ts, which uses JavaScript BigInt throughout — so the answers are exact regardless of length. The multiplicative formula multiplies before dividing at each step, which keeps every partial result a whole number and means the arithmetic never needs a rounding decision. The row sum and Pascal’s rule are computed from the entries rather than assumed. That engine is verified on every change against 39 hand-written assertions, including that the row of coefficients always sums to 2ⁿ and that Pascal’s rule holds at every tested position. The count and the per-case breakdown are published on the formula verification page.

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Read the guide

Binomial coefficients count subsets, and the Power Set Calculator lists the subsets those counts describe. For the same coefficients appearing as the terms of an expansion, the Square of a Binomial Calculator covers the case everyone meets first.