Create all possible words using a set or letters












2












$begingroup$


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











$endgroup$

















    2












    $begingroup$


    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











    $endgroup$















      2












      2








      2





      $begingroup$


      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 4 hours ago









      J. M. is slightly pensive♦

      98.3k10307466




      98.3k10307466










      asked 5 hours ago









      mf67mf67

      1026




      1026






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
            $endgroup$
            – mf67
            3 hours ago



















          3












          $begingroup$

          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|improve this answer









            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              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









              $endgroup$













              • $begingroup$
                Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
                $endgroup$
                – mf67
                3 hours ago
















              4












              $begingroup$

              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









              $endgroup$













              • $begingroup$
                Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
                $endgroup$
                – mf67
                3 hours ago














              4












              4








              4





              $begingroup$

              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 4 hours ago









              LeeLee

              47027




              47027












              • $begingroup$
                Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
                $endgroup$
                – mf67
                3 hours ago


















              • $begingroup$
                Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
                $endgroup$
                – mf67
                3 hours ago
















              $begingroup$
              Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
              $endgroup$
              – mf67
              3 hours ago




              $begingroup$
              Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
              $endgroup$
              – mf67
              3 hours ago











              3












              $begingroup$

              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









              $endgroup$


















                3












                $begingroup$

                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









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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 4 hours ago









                  bill sbill s

                  54.6k377157




                  54.6k377157























                      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|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|improve this answer









                        $endgroup$
















                          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|improve this answer









                          $endgroup$



                          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|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 3 hours ago









                          JagraJagra

                          7,85312159




                          7,85312159






























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