Get all permutations of a list in Haskell

I'm trying to do this from scratch, without the use of a library outside the standard lib. Heres my code: ``` permutations :: [a] -> [[a]] permutations (x:xs) = [x] : permutations' xs where permutations' (x:xs) = (:) <$> [x] <*> split xs split l = [[x] | x <- l] ``` The problem is that this only produces one fork of the non-deterministic computation. Ideally I'd want ``` (:) <$> [x] <*> ((:) <$> [x] <*> ((:) <$> [x] <*> ((:) <$> [x] <*> xs))) ``` But I can't find a way to do this cleanly. My desired result is something like this: ``` permutations "abc" -> ["abc", "acb", "bac", "bca", "cab", "cba"] ``` How do I do this?