Factorial of 252
252! = 204484624215022882284756162044645354527365854830301974733140854399740327680355894222586710783048275671670129947103938011481584892698594287523178247211816724109883700996451372905507541113381204196195195435460255293477573752210631662539582932406086262252828746817937545830842327002089213302615630504650695790421117659177816706362043873270049714165645369554463748888564282896170368904334765715609531606271737867415895310398229305913612697600000000000000000000000000000000000000000000000000000000000000
498 digits
Sumwise · Maintained by Theo Abbott (pen name) · automated checks · Data: exact arithmetic; unit definitions from NIST SP 811 · Updated · How the checks work
How it is calculated
252! = 252 × 251 × … × 2 × 1 = 2044846242150228822847561620446453545273…
Facts about 252!
| Digits | 498 |
|---|---|
| Trailing zeros | 62 (one for every factor 5 × 2) |
| Stirling estimate | ≈ Infinity |
| 252! ÷ 251! | 252 |
Nearby factorials
| 250! | 323285626090910773232081… (493 digits) |
| 251! | 811446921488186040812524… (495 digits) |
| 253! | 517346099264007892180433… (500 digits) |
| 254! | 131405909213058004613830… (503 digits) |
What are factorials used for?
Counting arrangements: 252! is the number of different orders in which 252 distinct items can be lined up. Factorials also appear in probability, combinations and series.
Counting with 252!
There are 2.0448 × 10^497 different ways to arrange 252 different books on a shelf: 252 choices for the first place, 251 for the second, and so on down to one choice for the last. If you only fill the first three places, as with gold, silver and bronze among 252 runners, there are 252 × 251 × 250 = 15,813,000 possible podiums. If the order does not matter, divide by 3! = 6 to get 2,635,500 ways to choose a group of three. Choosing a pair from 252 people can be done in 31,626 ways.
In probability, if 252 names are drawn out of a hat one at a time, the chance of drawing them in one particular order is 1 in 2.0448 × 10^497. That is why shuffled decks and passwords with many symbols are so hard to guess.
How big is 252!?
252! has 498 digits, so it is larger than the estimated number of atoms in the observable universe (about 10^80). The digits add up to 1917, a multiple of 9, as they are for every factorial from 6! on. Stirling's formula, √(2πn) × (n/e)^n, gives a close estimate without multiplying everything out, and gets more accurate as n grows.
Prime factors of 252!
Every prime up to 252 divides 252!. Legendre's formula counts how often: add ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … for each prime p. For 252! that gives 2^246 × 3^125 × 5^62 × 7^41 × 11^24 × 13^20 × 17^14 × 19^13 × 23^10 × 29^8 × … (54 primes in all). The 62 trailing zeros come from pairing the 62 factors of 5 with factors of 2.
Shuffles where nothing stays put
A related count is the number of derangements of 252 items: orders in which no item ends up in its original place, as in a Secret Santa where nobody draws their own name. The count is the whole number nearest to 252! ÷ e, so about 36.8% of all 2.0448 × 10^497 orders are derangements. From about four items on, that share stays close to 1 ÷ e, roughly 37%, however many items there are.
Computing 252! yourself
In Excel or Google Sheets, type =FACT(252). Spreadsheets keep only 15 significant digits, so from 18! on the result is rounded, and FACT stops at 170!, the largest factorial that fits in a standard floating-point number. In Python, math.factorial(252) returns the exact integer with all 498 digits, because Python integers have no size limit. The gamma function extends factorials to fractions: Γ(n + 1) = n!, so (1/2)! = √π ÷ 2 ≈ 0.8862.
Why is 0! equal to 1?
There is exactly one way to arrange nothing, and the rule n! = n × (n − 1)! only works for n = 1 if 0! = 1. Defining it this way keeps formulas for combinations and series correct at the edges.
How answers are calculated and checked: methodology.
Questions
What is 252 factorial?
252! is the product of every whole number from 1 to 252.
How many digits does 252! have?
498. The number of digits of n! is the whole part of log10(n!) plus one.
How many zeros are at the end of 252!?
62. Count the factors of 5 in the numbers 1 to 252: there are always more factors of 2, so every 5 makes one trailing zero.
Is 252! divisible by 253?
Yes. 253 is not prime, so its factors already appear among the numbers 1 to 252.
What is the last non-zero digit of 252!?
6. Strip the 62 trailing zeros and look at the digit that is left at the end.
What is 253! in terms of 252!?
253! = 253 × 252!. Each factorial is the previous one multiplied by the next whole number.