scramble prime challenge

[Albert Chan]

(02-18-2020 05:26 PM)Don Shepherd Wrote:  Yeah, I was hoping that, for example, 917371993771911 might be a scramble prime, but with
over 1 trillion permutations of those 15 digits, it would be highly unlikely.

With repeating digits, permutations should use multinomial coefficient

For 917371993771911, permutations = \(\large\binom{15}{5,2,4,4} = {15! \over 5! 2! 4! 4!} \) = 9,459,450

I made the same mistake with Python's itertools.permutations, which do n! permutations.
With repeated digits, many permutations produce the same patterns, wasting cycles.

Then, I googled and found this: Distinct permutations of the string | Set 2
Post #9 unique_permute were "borrowed" from article's Python3 code