Combination Calculator (nCr)

Exact value: Every digit of C(n, r), grouped with thousands separators.

Digits: How long the exact answer is — a quick sense of scale for large results.

Scientific notation: A 10-significant-figure approximation, shown when the exact value has 30 or more digits.

A combination is a selection where only membership matters: picking Alice and Bob is the same outcome as picking Bob and Alice. The count of such selections is the binomial coefficient, read as "n choose r":

The factorial form is the textbook definition, but it is a poor recipe for direct computation because n! explodes long before the final answer does. The multiplicative form cancels the (n − r)! factor first, which is exactly how this calculator works.

Standard JavaScript numbers hold integers exactly only up to 2^53 − 1, which is 9,007,199,254,740,991 — a 16-digit ceiling. Computing combinations through raw factorials fails even sooner: 171! overflows the floating-point range entirely, and precision quietly degrades long before that. Our general-purpose probability calculator takes that simpler factorial approach, which is fine for classroom-sized inputs but cannot show every digit of a large result.

This page instead runs a big-integer loop: starting from 1, it multiplies by (n − k + i) and divides by i for i = 1 up to k, where k = min(r, n − r) exploits the symmetry rule to halve the work. After step i the running value equals C(n − k + i, i), which is itself a binomial coefficient and therefore a whole number — so every intermediate division is exact and no rounding can creep in. That is how the tool reports all 30 digits of C(100, 50) = 100,891,344,545,564,193,334,812,497,256 rather than an estimate. If you care about the order of the chosen items rather than just membership, switch to the permutation calculator.

Combinations answer "how many groups are possible?" — the question behind committees, lottery tickets, and card hands alike:

• Committees: choosing 4 members from a 12-person board gives C(12, 4) = 495 distinct committees; a small 2-of-5 working group gives C(5, 2) = 10.
• Lotteries: a 6-of-49 draw has C(49, 6) = 13,983,816 equally likely tickets, so one ticket wins the jackpot with probability 1 in 13,983,816.
• Card hands: a five-card poker hand from a 52-card deck is one of C(52, 5) = 2,598,960 possibilities.

In every case the division by r! is doing the conceptual work: it collapses all the orderings of the same group into a single outcome.

Binomial coefficients are the entries of Pascal's triangle: row n lists C(n, 0) through C(n, n), and each entry is the sum of the two entries above it.

Two consequences are worth remembering. Row sums equal 2^n because every subset of an n-element set is either chosen or not, element by element. And the rows read the same in both directions — the symmetry C(n, r) = C(n, n − r) — because choosing r items to keep is the same decision as choosing n − r items to leave out. Try n = 49 with r = 43: the calculator returns 13,983,816, identical to C(49, 6), and it exploits that same symmetry internally to keep large computations fast.

How many different tickets exist in a lottery where players pick 6 numbers out of 49? Enter n = 49 and r = 6. By hand, the multiplicative form keeps the numbers manageable:

1. Cancel the (n − r)! factor: C(49, 6) = (49 × 48 × 47 × 46 × 45 × 44) / 6!.
2. Build the numerator: 49 × 48 = 2,352; × 47 = 110,544; × 46 = 5,085,024; × 45 = 228,826,080; × 44 = 10,068,347,520.
3. Divide by 6! = 720: 10,068,347,520 ÷ 720 = 13,983,816.

The calculator's interleaved loop reaches the same answer while keeping every intermediate value small and whole: 44, then 990, 15,180, 178,365, 1,712,304, and finally 13,983,816 — each partial result is itself a binomial coefficient.

As a probability, one ticket hits the jackpot with chance 1 ÷ 13,983,816 ≈ 0.0000000715, or about 7.15 × 10^-8. Matching exactly 5 of the 6 winning numbers can happen in C(6, 5) × C(43, 1) = 6 × 43 = 258 ways, so those odds are 13,983,816 ÷ 258 ≈ 54,200.84 — about 1 in 54,201.