For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other. You're only looking for the millionth lexicographic permutation of "0123456789" Problem 24 A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are: 012 021 102 120 201 210 What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? <Spoiler> <Spoiler> <Spoiler> Plan of attack: -- The "x"s are different numbers -- 0xxxxxxxxx represents 9! = 362880 permutations/numbers -- 1xxxxxxxxx represents 9! = 362880 permutations/numbers -- 2xxxxxxxxx represents 9! = 362880 permutations/numbers -- 20xxxxxxxx represents 8! = 40320 -- 21xxxxxxxx represents 8! = 40320 -- 23xxxxxxxx represents 8! = 40320 -- 24xxxxxxxx represents 8! = 40320 -- 25xxxxxxxx represents 8! = 40320 -- 26xxxxxxxx represents 8! = 40320 -- 27xxxxxxxx represents 8! = 40320 etc. <Spoiler> <Spoiler> -- lexOrder "0123456789" 1000000 "" lexOrder digits left s | len == 0 = s ++ digits | quot > 0 && rem == 0 = lexOrder (digits\\(show (digits!!(quot-1)))) rem (s ++ [(digits!!(quot-1))]) | quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len))) rem (s ++ [(digits!!len)]) | rem == 0 = lexOrder (digits\\(show (digits!!(quot+1)))) rem (s ++ [(digits!!(quot+1))]) | otherwise = lexOrder (digits\\(show (digits!!(quot)))) rem (s ++ [(digits!!(quot))]) where len = (length digits) - 1 facLen = factorial len (quot,rem) = quotRem left facLen