1

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}

Please help me :)

J. M.'s missing motivation
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Andrej
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1 Answers1

3

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]
Simon Woods
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