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10! = 3,628,800 ways to line up 10 people. C(52,5) = 2,598,960 poker hands. Computed with exact integers — every digit, no rounding.
The counting questions that run statistics, probability, and card games: how many ways to arrange, select, or order n things. This calculator computes factorials, permutations, and combinations with exact integer arithmetic — 20! is 2,432,902,008,176,640,000 to the last digit, not the 2.43e18 approximation a normal calculator rounds to.
Stats students use it for the counting questions: 10! = 3,628,800 arrangements, P(10,3) = 720 podiums, C(52,5) = 2,598,960 poker hands — exact BigInt digits, no float lies.
Order. A permutation counts arrangements where sequence matters; a combination counts selections where it does not. Picking 3 people from 10 for gold-silver-bronze is a permutation: P(10,3) = 10×9×8 = 720 — because Alice-Bob-Carol and Carol-Alice-Bob are different podiums. Picking 3 people from 10 for a committee is a combination: C(10,3) = 720 ÷ 3! = 120 — the 6 orderings of the same trio collapse into one selection. The mnemonic: permutations are for podiums and passwords; combinations are for committees and cards. Poker hands are the famous case: C(52,5) = 2,598,960 possible 5-card hands, which is why a royal flush (4 of those hands) shows up once per 649,740 deals.
Each step multiplies by the next integer, and multiplication compounds brutally: 10! = 3.6 million, 13! = 6.2 billion, 20! = 2.4 quintillion, and 52! — the ways to shuffle a card deck — is an 68-digit number larger than the count of atoms on Earth. The famous consequence: every properly shuffled deck in history has almost certainly produced an order that has never existed before. This explosive growth is why brute-force approaches die in computing: a traveling-salesman route through 15 cities has 1.3 trillion orderings (15!÷2), and 25 cities has more than any supercomputer can enumerate. It is also why combinations shrink things so much — dividing by k! removes the ordering explosion.
Zero factorial equals 1 — by definition, but a sensible one. Two ways to see it. First, the pattern: n! = (n+1)! ÷ (n+1), so 4! = 5!÷5 = 24, 3! = 24÷4 = 6, 2! = 2, 1! = 1, and 0! = 1!÷1 = 1 — the recursion breaks anywhere else. Second, the meaning: n! counts arrangements of n items, and there is exactly one way to arrange zero items — the empty arrangement. It also has to be 1 for the formulas to work: C(n,n) = n! ÷ (n! × 0!) must equal 1 (there is exactly one way to choose everything), which forces 0! = 1. Same logic as x⁰ = 1: empty products equal the multiplicative identity.