Something like the following: Map[Sort[#] &, list] (*{{{a, 1}, {b, 3}, {c, 5}}, {{a, 5}, {b, 1}, {c, 3}}, {{a, 5}, {b, 3}, {c, 1}}, {{a, 1}, {b, 5}, {c, 3}}}*) A first approximation: MyOrderList[list_List, order_?(Positive[#] && Element[#, Integers] &)] := Block[{slist, perm, mylist}, slist := Map[Sort, list]; perm = Mean[Map[Composition[Length, Permutations[#] &], slist]]; mylist := Table[Table[Extract[Select[Tuples[slist[[i]], Length[slist[[i]]]], ContainsAll[#, slist[[i]]] &], j], {i, 1, Length[slist]}], {j, 1, perm}][[order]]; Return[If[order <= perm, mylist, HoldForm[MyOrderList]]]; ]; Tests: MyOrderList[list, 1] (*{{{a, 1}, {b, 3}, {c, 5}}, {{a, 5}, {b, 1}, {c, 3}}, {{a, 5}, {b, 3}, {c, 1}}, {{a, 1}, {b, 5}, {c, 3}}}*) MyOrderList[list, 2] (*{{{a, 1}, {c, 5}, {b, 3}}, {{a, 5}, {c, 3}, {b, 1}}, {{a, 5}, {c, 1}, {b, 3}}, {{a, 1}, {c, 3}, {b, 5}}}*) MyOrderList[list, 3] (*{{{b, 3}, {a, 1}, {c, 5}}, {{b, 1}, {a, 5}, {c, 3}}, {{b, 3}, {a, 5}, {c, 1}}, {{b, 5}, {a, 1}, {c, 3}}}*) MyOrderList[list,4] (*{{{b, 3}, {c, 5}, {a, 1}}, {{b, 1}, {c, 3}, {a, 5}}, {{b, 3}, {c, 1}, {a, 5}}, {{b, 5}, {c, 3}, {a, 1}}}*) MyOrderList[list,5] (*{{{c, 5}, {a, 1}, {b, 3}}, {{c, 3}, {a, 5}, {b, 1}}, {{c, 1}, {a, 5}, {b, 3}}, {{c, 3}, {a, 1}, {b, 5}}}*) MyOrderList[list,6] (*{{{c, 5}, {b, 3}, {a, 1}}, {{c, 3}, {b, 1}, {a, 5}}, {{c, 1}, {b, 3}, {a, 5}}, {{c, 3}, {b, 5}, {a, 1}}}*) Your problem in general is somewhat complicated for more characters, but I will think a little more about it.