                    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)


permute = function(v, m){ //v1.0
    for(var p = -1, j, k, f, r, l = v.length, q = 1, i = l + 1; --i; q *= i);
    for(x = [new Array(l), new Array(l), new Array(l), new Array(l)], j = q, k = l + 1, i = -1;
        ++i < l; x[2][i] = i, x[1][i] = x[0][i] = j /= --k);
    for(r = new Array(q); ++p < q;)
        for(r[p] = new Array(l), i = -1; ++i < l; !--x[1][i] && (x[1][i] = x[0][i],
            x[2][i] = (x[2][i] + 1) % l), r[p][i] = m ? x[3][i] : v[x[3][i]])
            for(x[3][i] = x[2][i], f = 0; !f; f = !f)
                for(j = i; j; x[3][--j] == x[2][i] && (x[3][i] = x[2][i] = (x[2][i] + 1) % l, f = 1));
    return r;
};
 
list = [ 'ABCD', 'CABD', 'ACDB', 'DACB', 'BCDA', 'ACBD', 'ADCB', 'CDAB',
        'DABC', 'BCAD', 'CADB', 'CDBA', 'CBAD', 'ABDC', 'ADBC', 'BDCA',
        'DCBA', 'BACD', 'BADC', 'BDAC', 'CBDA', 'DBCA', 'DCAB'];
 
all = permute(list[0].split('')).map(function(elem) {return elem.join('')});
 
missing = all.filter(function(elem) {return list.indexOf(elem) == -1});
print(missing);  // ==> DBAC