Create all possible words using a set or letters












1












$begingroup$


Given a list of letters,



letters = { "A", "B", ..., "F" }


is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.






string-manipulation combinatorics







share|improve this question











$endgroup$

















    1












    $begingroup$


    Given a list of letters,



    letters = { "A", "B", ..., "F" }


    is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.






    string-manipulation combinatorics







    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given a list of letters,



      letters = { "A", "B", ..., "F" }


      is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.






      string-manipulation combinatorics







      share|improve this question











      $endgroup$




      Given a list of letters,



      letters = { "A", "B", ..., "F" }


      is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.






      string-manipulation combinatorics




      string-manipulation combinatorics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      J. M. is slightly pensive♦

      98.3k10306466




      98.3k10306466










      asked 1 hour ago









      mf67mf67

      975




      975






















          2 Answers
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          active

          oldest

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          2












          $begingroup$

          You can create permutations with all of the letters as strings with:



          StringJoin /@ Permutations[letters]


          If you want lists of the individual letters just use:



          Permutations[letters]


          Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.






          share|improve this answer









          $endgroup$





















            1












            $begingroup$

            Pemutations will do it:



            letters = {"a", "b", "c"};
            
Permutations[letters, {3}]
            
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"}, 
            
 {"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}





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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

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              active

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              2












              $begingroup$

              You can create permutations with all of the letters as strings with:



              StringJoin /@ Permutations[letters]


              If you want lists of the individual letters just use:



              Permutations[letters]


              Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.






              share|improve this answer









              $endgroup$


















                2












                $begingroup$

                You can create permutations with all of the letters as strings with:



                StringJoin /@ Permutations[letters]


                If you want lists of the individual letters just use:



                Permutations[letters]


                Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.






                share|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You can create permutations with all of the letters as strings with:



                  StringJoin /@ Permutations[letters]


                  If you want lists of the individual letters just use:



                  Permutations[letters]


                  Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.






                  share|improve this answer









                  $endgroup$



                  You can create permutations with all of the letters as strings with:



                  StringJoin /@ Permutations[letters]


                  If you want lists of the individual letters just use:



                  Permutations[letters]


                  Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  LeeLee

                  45027




                  45027























                      1












                      $begingroup$

                      Pemutations will do it:



                      letters = {"a", "b", "c"};
                      
Permutations[letters, {3}]
                      
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"}, 
                      
 {"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}





                      share|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Pemutations will do it:



                        letters = {"a", "b", "c"};
                        
Permutations[letters, {3}]
                        
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"}, 
                        
 {"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}





                        share|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Pemutations will do it:



                          letters = {"a", "b", "c"};
                          
Permutations[letters, {3}]
                          
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"}, 
                          
 {"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}





                          share|improve this answer









                          $endgroup$



                          Pemutations will do it:



                          letters = {"a", "b", "c"};
                          
Permutations[letters, {3}]
                          
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"}, 
                          
 {"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 1 hour ago









                          bill sbill s

                          54.6k377156




                          54.6k377156






























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