# Digits in factorial numbers

Several of my students have noticed that the factorial of 23 is the first to contain at least one copy of each of the ten digits, and they wonder whether every sufficiently large factorial similarly contains at least one of each of the ten digits.

I would guess yes, but has this been proved?

• $24!$ does not have any digit $5$ appearing. "Arbitrarily large factorial"... your question seems to be then... does there exist some $N$ such that for all $n\geq N$ you have $n!$ has all ten digits appearing in its base-10 representation... noting that $N$ would not be $23$. Commented Apr 5, 2021 at 17:19
• Clarified the issue. Commented Apr 5, 2021 at 17:32
• $42!$ through $999!$ all have all $10$ digits, so perhaps $42$ is a good candidate. In python max(n for n in range(25,1000) if len(set(str(factorial(n)))) != 10) evaluates to $41$ Commented Apr 5, 2021 at 17:39
• Though this doesn't answer the question: in case digits are uniformly distributed in the factorials, one notes that given sufficient digits (or given a large enough factorial), the probability of not finding all 10 digits can be made arbitrarily small Commented Apr 5, 2021 at 17:40
• Commented Apr 5, 2021 at 17:54