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






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          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












            $begingroup$

            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









            $endgroup$





















              0












              $begingroup$

              If I follow the OP's question, I think they want the following:



              letters = {"a", "b", "c"};
              
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
              
(StringJoin[#] & /@ #) & /@ p
              

              
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}




              share









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                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









                $endgroup$


















                  3












                  $begingroup$

                  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









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    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

                    46027




                    46027























                        2












                        $begingroup$

                        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









                        $endgroup$


















                          2












                          $begingroup$

                          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









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            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























                                0












                                $begingroup$

                                If I follow the OP's question, I think they want the following:



                                letters = {"a", "b", "c"};
                                
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
                                
(StringJoin[#] & /@ #) & /@ p
                                

                                
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}




                                share









                                $endgroup$


















                                  0












                                  $begingroup$

                                  If I follow the OP's question, I think they want the following:



                                  letters = {"a", "b", "c"};
                                  
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
                                  
(StringJoin[#] & /@ #) & /@ p
                                  

                                  
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}




                                  share









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    If I follow the OP's question, I think they want the following:



                                    letters = {"a", "b", "c"};
                                    
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
                                    
(StringJoin[#] & /@ #) & /@ p
                                    

                                    
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}




                                    share









                                    $endgroup$



                                    If I follow the OP's question, I think they want the following:



                                    letters = {"a", "b", "c"};
                                    
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
                                    
(StringJoin[#] & /@ #) & /@ p
                                    

                                    
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}





                                    share











                                    share


                                    share










                                    answered 18 secs ago









                                    JagraJagra

                                    7,85312159




                                    7,85312159






























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