Factorial of 433

Counting with 433!

There are 1.8499 × 10^955 different ways to arrange 433 different books on a shelf: 433 choices for the first place, 432 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 433 runners, there are 433 × 432 × 431 = 80,621,136 possible podiums. If the order does not matter, divide by 3! = 6 to get 13,436,856 ways to choose a group of three. Choosing a pair from 433 people can be done in 93,528 ways.

In probability, if 433 names are drawn out of a hat one at a time, the chance of drawing them in one particular order is 1 in 1.8499 × 10^955. That is why shuffled decks and passwords with many symbols are so hard to guess.

How big is 433!?

433! has 956 digits, so it is larger than the estimated number of atoms in the observable universe (about 10^80). The digits add up to 3771, 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 433!

Every prime up to 433 divides 433!. Legendre's formula counts how often: add ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … for each prime p. For 433! that gives 2^428 × 3^214 × 5^106 × 7^70 × 11^42 × 13^35 × 17^26 × 19^23 × 23^18 × 29^14 × … (84 primes in all). The 106 trailing zeros come from pairing the 106 factors of 5 with factors of 2.

Shuffles where nothing stays put

A related count is the number of derangements of 433 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 433! ÷ e, so about 36.8% of all 1.8499 × 10^955 orders are derangements. From about four items on, that share stays close to 1 ÷ e, roughly 37%, however many items there are.

Computing 433! yourself

In Excel or Google Sheets, type =FACT(433). 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(433) returns the exact integer with all 956 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.