(**) Generate the combinations of K distinct objects chosen from the N elements of a list
In how many ways can a committee of 3 be chosen from a group of 12 people? We all know that there are C(12,3) = 220 possibilities (C(N,K) denotes the well-known binomial coefficients). For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.
-- Import the 'tails' function -- > tails [0,1,2,3] -- [[0,1,2,3],[1,2,3],[2,3],,] import Data.List (tails) -- The implementation first checks if there's no more elements to select, -- if so, there's only one possible combination, the empty one, -- otherwise we need to select 'n' elements. Since we don't want to -- select an element twice, and we want to select elements in order, to -- avoid combinations which only differ in ordering, we skip some -- unspecified initial elements with 'tails', and select the next element, -- also recursively selecting the next 'n-1' element from the rest of the -- tail, finally consing them together -- Using list comprehensions combinations :: Int -> [a] -> [[a]] combinations 0 _ = [  ] combinations n xs = [ y:ys | y:xs' <- tails xs , ys <- combinations (n-1) xs'] -- Alternate syntax, using 'do'-notation combinations :: Int -> [a] -> [[a]] combinations 0 _ = return  combinations n xs = do y:xs' <- tails xs ys <- combinations (n-1) xs' return (y:ys)
combinations :: Int -> [a] -> [[a]] -- We don't actually need this base case; it's just here to -- avoid the warning about non-exhaustive pattern matches combinations _  = [] -- Base case--choosing 0 elements from any list gives an empty list combinations 0 _ = [] -- Get all combinations that start with x, recursively choosing (k-1) from the -- remaining xs. After exhausting all the possibilities starting with x, if there -- are at least k elements in the remaining xs, recursively get combinations of k -- from the remaining xs. combinations k (x:xs) = x_start ++ others where x_start = [ x : rest | rest <- combinations (k-1) xs ] others = if k <= length xs then combinations k xs else 
Variant: This version avoids getting a result consisting of the empty choice if we ask for, say, 10 elements from the empty list. (Instead, an empty list, suggesting "no options available" is returned.)
combinations :: Int -> [a] -> [[a]] combinations 0 _ = [] combinations _  =  combinations n (x:xs) = (map (x:) (combinations (n-1) xs)) ++ (combinations n xs)
A solution using subsequences in Data.List
combinations k ns = filter ((k==).length) (subsequences ns)