                    ABCD
                    CABD
                    ACDB
                    DACB
                    BCDA
                    ACBD
                    ADCB
                    CDAB
                    DABC
                    BCAD
                    CADB
                    CDBA
                    CBAD
                    ABDC
                    ADBC
                    BDCA
                    DCBA
                    BACD
                    BADC
                    BDAC
                    CBDA
                    DBCA
                    DCAB

Listed above are   all-but-one   of the permutations of the symbols   A,   B,   C,   and   D,   except   for one permutation that's   not   listed.



Task

Find that missing permutation.



Methods

 Obvious method: 
        enumerate all permutations of   A,  B,  C,  and  D,  
        and then look for the missing permutation. 

 alternate method:
        Hint:  if all permutations were shown above,  how many 
        times would  A  appear in each position?     
        What is the  parity  of this number?

 another alternate method:
        Hint:  if you add up the letter values of each column, 
        does a missing letter   A,  B,  C,  and  D   from each
        column cause the total value for each column to be unique?



Related task

   Permutations)


(* insert x at all positions into li and return the list of results *)
let rec insert x = function
  | [] -> [[x]]
  | a::m as li -> (x::li) :: (List.map (fun y -> a::y) (insert x m))
 
(* list of all permutations of li *)
let permutations li = 
  List.fold_right (fun a z -> List.concat (List.map (insert a) z)) li [[]]
 
(* convert a string to a char list *)
let chars_of_string s =
  let cl = ref [] in
  String.iter (fun c -> cl := c :: !cl) s;
  (List.rev !cl)
 
(* convert a char list to a string *)
let string_of_chars cl =
  String.concat "" (List.map (String.make 1) cl)