Find next lexicographic - permutation (permutations as it would occur in a dictionary order).

E.g: nextP('ABCD') = ABDC

If you can generate the next permutation, then you can also generate a 'cycle' of all permutations using a construct like,

     cycle = +1;
     curr = start;
     while ( true )         
         curr = nextP(curr);
         if ( curr == start )
             break;
         end
         cycle = cycle+1;
     end

For fun, you could generate all the n! permutations of a, unique n-letter string.