# 99 questions/Solutions/26

### From HaskellWiki

(**) 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],[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)