# Descending order of Permutations using Select[Tuples[{5, 4, 3, 2, 1}, {3}], OrderedQ]

I was wondering if there is a possible way to still use OrderedQ and get a descending order of permutations: This is Ascending order:

Select[Tuples[{1, 2, 3, 4, 5}, {3}], OrderedQ]
{{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2},
{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5},
{1, 4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4},
{2, 2, 5}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 4, 4}, {2, 4, 5},
{2, 5, 5}, {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 4}, {3, 4, 5},
{3, 5, 5}, {4, 4, 4}, {4, 4, 5}, {4, 5, 5}, {5, 5, 5}


And i need them in Descending order, for eg:

{{3,2,1},{3,1,1},{1,1,1}.. etc}


• What if you just replace 1->6, 2->5,3->4,4->3,5->2,6->1? May 23 '15 at 12:22
• Related: (82522), (83791), (83938), (84023), (84147) May 23 '15 at 12:24
• How about using Reverse on your output lists? Or using OrderedQ@Reverse@#& as your test function? May 23 '15 at 12:25
• ... or Select[Tuples[Reverse@{1, 2, 3, 4, 5}, {3}], GreaterEqual @@ # &]
– kglr
May 23 '15 at 13:51
• Ok thanks guys! May 23 '15 at 16:36

Some options:

t = Tuples[{1, 2, 3, 4, 5}, {3}];

Reverse /@ Select[t, OrderedQ]

Select[t, OrderedQ] ~Reverse~ 2

Select[t, OrderedQ @* Reverse]

Select[t, Reverse /* OrderedQ]

Select[t, OrderedQ[#, GreaterEqual] &]

6 - Select[t, OrderedQ]

• Wow, thanks so much! May 23 '15 at 16:17