Factorial (n) for n > 24 is greater than 10^n.

Reminds me of: Professor asked us to find the biggest factorial using C programming language. And then using LISP. You can imagine our surprise.

There is a algorithm call Prime Swing Factorial that can compute large factorials exactly in arbitrary precision math using prime factorization. Like 10000000! in under second depending of how optimized the math library it. Probably like 100x faster than the normal method.

The author's casual mention of 52! at the opening of the article triggered an OLD webpage that I saw many years ago

https://czep.net/weblog/52cards.html

Anyone know how to determine the age of this page (it's got be at least 20yrs old)

The main.css file it imports dates itself to March 9 of 2005, and is housed in an "ancient history" section of the website that covers everything before October 26, 2010, so: "sometime between those two years" =P

A quick and dirty approximation of the number of digits in n! is n lg n, which approximates n! from above, via the inequality

  1 * 2 * … * n ≤ n * … * n.

(This approximation should be familiar to many from an algorithmics class.)

For a tighter bound, use n lg n - n/2, or a better approximation of ln 10 in place of 1/2 if you wish. This comes from Stirling's approximation which notes that

  ln n! = n ln n - n + O(ln n).

(This approximation should be familiar to many from an algorithmics class.)

You need both sides though :)

What makes it interesting for estimating algorithmic complexity is that \log{n!} \in \Theta(n \log n). One side is obvious as you note, the other less so, but there's a famous trick to do both at once:

\log{n!} = \log{\prod_{h=0}^{n} h} = \sum_{h=0}^{n} \log{h}

Therefore,

\int_0^n \log{x} dx \le \log{n!} \le \int_0^n \log{x+1} dx

with both integrals trivial by parts.

lg(n!) grows roughly as (n lg n). Constants matter, of course, but to that's the rough estimate.

As an aside, if you take numbers from 0 to (n-1) in an array, there are n! configurations, so representing each configuration or differentiating each configuration take n lg n bits. So, in some sense, taking a mapping that's able to differentiate the input state to map to the ordered state takes at least O(n lg n) time, the standard runtime of a basic sorting algorithm.

Any additional assumptions (n larger than maximum element, distribution of elements) helps reduce this.

What's surprising (to many) is that

n! < exp(n log n)

What's surprising (to many) is that

n! < exp(n log n)

Why would that be surprising? I can see that many wouldn’t know whether it’s true, but

  exp(n × log n) =
  exp(log(n) × n) =
  exp(log(n))^n =
  n^n

and it’s not surprising that

    1 × 2 × 3 × 4 × … × n
  < n × n × n × n × … × n

for n > 1

My favorite one is with the 52! seconds:

Start a timer that will count down the number of seconds from 52! to 0. Then walk around the Earth’s equator with one step every billion years. Then, after you make your way around the earth equator (by taking 1 step every billion of years), you take one drop of water out of the Pacific Ocean. Then, you repeat the process of walking around the equator, and everytime you walk around, you keep draining one singular drop of water. After the ocean is fully drained, you refill the ocean and put a piece of paper underneath you. Now, you once again repeat this process of walking, draining, and placing papers. After your stack of papers has reached the Sun, you repeat another 1000 times.

After all this, you have completed just about a third of the timer.

https://sites.imsa.edu/hadron/2025/02/26/how-big-is-52/