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I've got a certain class of lists with, say, 4 elements, that are lists themselves. E.g. {{1, 2, 3}, {4}, {5, 6}, {7}}. I'd like to find a way to generate a list that contains all of the cousins of this list where the sublists are permuted in all possible ways. With the previous example, this list would have size $3!\times 2!=12$.

To be really definite, in the simpler case of {{1, 2},{3,4}}, I'd want to obtain {{{1, 2},{3,4}},{{2,1},{3,4}},{{1, 2},{4,3}},{{2,1},{4,3}}}.

I feel like there should be a one line code for this but I couldn't figure it out.

Permutations/@{{1, 2, 3}, {4}, {5, 6}, {7}} returns a reasonably interesting list, out of which I'm unable to extract what I want in a simple (i.e. Mathematica) way. Using further Pick starts to demand that I get more explicitly certain lists or binary trees or so...

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    $\begingroup$ Does Tuples[Permutations /@ {{1, 2, 3}, {4}, {5, 6}, {7}}] suit your needs? $\endgroup$
    – J. M.'s lack of A.I. ♦
    Mar 31, 2016 at 17:28
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    $\begingroup$ @J.M. Oh... Tuples is so much simpler than Outer for this... If you post that as an answer, I'll delete mine. $\endgroup$
    – Martin Ender
    Mar 31, 2016 at 17:31
  • $\begingroup$ It's not too late to edit your answer, @Martin. ;) $\endgroup$
    – J. M.'s lack of A.I. ♦
    Mar 31, 2016 at 17:32
  • $\begingroup$ @J.M. Yes it does ! Please put this as an answer. I also like very much the explanation with Outer below. $\endgroup$
    – picop
    Mar 31, 2016 at 17:33
  • $\begingroup$ @Martin Büttner well it is simpler yes but I'm also happy that I learned about Outer today as well ! $\endgroup$
    – picop
    Mar 31, 2016 at 17:34

1 Answer 1

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What you're looking for is the outer product of all the lists returned by Permutations /@ .... You can use Outer for that. The only issue is that Outer returns the result as a nested list (basically an n-dimensional table of all the possible permutations), so you'll have to flatten it afterwards:

lists = {{1, 2, 3}, {4}, {5, 6}, {7}};
Flatten[Outer[List, ##, 1] & @@ Permutations /@ lists, Length@lists - 1]

(* {{{1, 2, 3}, {4}, {5, 6}, {7}}, {{1, 2, 3}, {4}, {6, 5}, {7}}, 
    {{1, 3, 2}, {4}, {5, 6}, {7}}, {{1, 3, 2}, {4}, {6, 5}, {7}}, 
    {{2, 1, 3}, {4}, {5, 6}, {7}}, {{2, 1, 3}, {4}, {6, 5}, {7}}, 
    {{2, 3, 1}, {4}, {5, 6}, {7}}, {{2, 3, 1}, {4}, {6, 5}, {7}}, 
    {{3, 1, 2}, {4}, {5, 6}, {7}}, {{3, 1, 2}, {4}, {6, 5}, {7}}, 
    {{3, 2, 1}, {4}, {5, 6}, {7}}, {{3, 2, 1}, {4}, {6, 5}, {7}}} *)

Well if this is now the accepted answer, I might as well edit in J. M.'s solution for completeness. The above is essentially a manual reimplementation of the built-in Tuples so that's all you need:

Tuples[Permutations /@ lists]
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