How this instrument works
A factorial multiplies a whole number by every positive whole number smaller than it: n! = n × (n−1) × (n−2) × … × 2 × 1. The result counts something specific — the number of distinct ways to arrange n distinct objects in a row. Picture five chairs in a line: the first can be filled by any of 5 people, the second by any of the remaining 4, the third by any of the remaining 3, and so on down to a single person left for the last chair. Multiplying those shrinking choices together, 5 × 4 × 3 × 2 × 1, is exactly what n! computes.
The definition is recursive at heart: n! = n × (n−1)!, each factorial built from the one before it. Running that rule down to its floor raises the question of what 0! should be, and the answer is 1 — not because zero objects can be arranged in one visible way, but because a product with no factors in it is, by the same convention that makes an empty sum equal 0, defined as 1. That single convention keeps the recursive rule intact all the way to its base case and keeps combinatorial formulas built on factorials well-behaved at their edges.
What makes factorials worth a dedicated instrument is how sharply they accelerate. Going from 5! to 10! only doubles n, yet the result jumps from 120 to 3,628,800 — a factor of over thirty thousand. Unlike an exponential such as 2^n, which multiplies by the same fixed ratio at every step, n! multiplies by a ratio that itself grows with n, so it eventually outruns any exponential no matter how large that exponential's base is. That growth also has a real limit inside a computer: 170! is about as large a factorial as an ordinary double-precision number can hold, and 171! overflows it.
- Enter the number of objects to arrange into the n field — any whole number from 0 up to 170.
- Read the count of possible orderings in n!, the field that reports the product.
- Try n = 0 first to see the empty-product convention in action: n! returns exactly 1.
- Raise n by just one or two at a time and watch how quickly n! climbs — the jump between neighbouring values grows every step.
- Keep n at 170 or below; beyond that, standard floating-point arithmetic runs out of room to hold the exact result.
Worked example — seating five people in a row
Five distinct people need to be seated in five chairs lined up in a row, and every different seating order counts as a separate arrangement. Set n = 5: the first chair can be filled 5 ways, the second 4 ways with one person already seated, the third 3 ways, the fourth 2 ways, and the last chair only 1 way once everyone else is placed. Multiplying those counts together gives n! = 5 × 4 × 3 × 2 × 1 = 120, so there are 120 genuinely different ways to seat the group.
Push the same idea five seats further and the count does not merely double: 10! works out to 3,628,800, more than thirty thousand times the five-seat total, even though n only doubled from 5 to 10. That is the defining shape of factorial growth — each additional person multiplies the running total by a bigger factor than the person before, which is why factorials show up wherever the question is precisely 'how many complete orderings exist,' from shuffled decks of cards to scheduling problems with more entries than seats at this table.
Questions
What does 5 factorial actually count?
5! = 120 counts every possible ordering of 5 distinct objects — every way to seat 5 people in a row, shuffle 5 cards, or sequence 5 tasks. The multiplication 5 × 4 × 3 × 2 × 1 mirrors filling five slots one at a time: five choices for the first slot, four remaining for the second, and so on down to a single choice left for the last.
Why is 0! defined as 1 instead of 0?
Because a factorial is a product, and the product of zero factors — the empty product — is defined as 1, the multiplicative identity, the same way a sum of zero terms is defined as 0. That convention keeps the recursive rule n! = n × (n−1)! consistent down to n = 0, and it is what lets formulas built from factorials, such as counting combinations, behave correctly at their boundary instead of needing a special-case rule bolted on.
How fast does factorial growth actually accelerate?
Far faster than exponential growth. 10! is already 3,628,800, and 20! runs to 19 digits — over 2.4 quintillion. An exponential like 2^n multiplies by the same fixed ratio every step, but n! multiplies by a ratio that itself keeps climbing, one step larger each time, which is why factorials eventually overtake any fixed-base exponential no matter how large that base is set.
Why does this calculator cap out at n = 170?
Because 171! works out to roughly 1.24 × 10^309, just past the roughly 1.80 × 10^308 ceiling a standard double-precision number can hold, so it overflows to infinity in floating-point arithmetic. 170! itself is a comparatively modest 7.26 × 10^306 and fits with room to spare — it isn't a flaw in the formula, n! is perfectly well defined for any whole number, it's simply where ordinary computer arithmetic runs out of digits.
Where does the factorial symbol come from?
It's a single exclamation mark placed right after the number, so five factorial is written 5!. The French mathematician Christian Kramp introduced the notation in 1808 as compact shorthand for a product that had previously been spelled out in full, and it has stayed essentially unchanged in the two centuries since.
Is there a shortcut for estimating a very large factorial?
Stirling's approximation, n! ≈ √(2πn) × (n/e)^n, estimates factorials for large n without multiplying out the full product, and its relative error shrinks as n grows. It's the standard tool in combinatorics and statistics for gauging how something like 100! behaves without ever computing all 158 digits of the exact value.