List all possible 4 chooses from 9 in Haskell

Question

I\'m not able to find an effective way to pick out all permutations of 4 elements from a list of 9 elements in Haskell. The python-way to do the same thing:

Answer 1

pick :: Int -> [a] -> [[a]]
pick 0 _ = [[]]
pick _ [] = []
pick n (x : xs) = map (x :) (pick (n - 1) xs) ++ pick n xs

perms :: Int -> [a] -> [[a]]
perms n l = pick n l >>= permutations