Create all possible words using a set or letters












4












$begingroup$


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











$endgroup$

















    4












    $begingroup$


    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











    $endgroup$















      4












      4








      4





      $begingroup$


      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 yesterday









      J. M. is slightly pensive♦

      98.5k10308466




      98.5k10308466










      asked yesterday









      mf67mf67

      1126




      1126






















          3 Answers
          3






          active

          oldest

          votes


















          9












          $begingroup$

          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"}}


          To get all six-letter words:



          letters = {"a", "b", "c", "d", "e", "f"};
          
perms = Permutations[letters, {6}];
          
StringJoin /@ perms
          

          
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.


          there are a lot of them.






          share|improve this answer











          $endgroup$





















            7












            $begingroup$

            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
              yesterday



















            3












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              No, the OP requested all six letter words.
              $endgroup$
              – m_goldberg
              yesterday










            • $begingroup$
              Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
              $endgroup$
              – Doorknob
              yesterday











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9












            $begingroup$

            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"}}


            To get all six-letter words:



            letters = {"a", "b", "c", "d", "e", "f"};
            
perms = Permutations[letters, {6}];
            
StringJoin /@ perms
            

            
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.


            there are a lot of them.






            share|improve this answer











            $endgroup$


















              9












              $begingroup$

              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"}}


              To get all six-letter words:



              letters = {"a", "b", "c", "d", "e", "f"};
              
perms = Permutations[letters, {6}];
              
StringJoin /@ perms
              

              
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.


              there are a lot of them.






              share|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                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"}}


                To get all six-letter words:



                letters = {"a", "b", "c", "d", "e", "f"};
                
perms = Permutations[letters, {6}];
                
StringJoin /@ perms
                

                
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.


                there are a lot of them.






                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"}}


                To get all six-letter words:



                letters = {"a", "b", "c", "d", "e", "f"};
                
perms = Permutations[letters, {6}];
                
StringJoin /@ perms
                

                
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.


                there are a lot of them.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited yesterday

























                answered yesterday









                bill sbill s

                54.7k377157




                54.7k377157























                    7












                    $begingroup$

                    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
                      yesterday
















                    7












                    $begingroup$

                    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
                      yesterday














                    7












                    7








                    7





                    $begingroup$

                    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 yesterday









                    LeeLee

                    50027




                    50027












                    • $begingroup$
                      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
                      yesterday


















                    • $begingroup$
                      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
                      yesterday
















                    $begingroup$
                    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
                    yesterday




                    $begingroup$
                    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
                    yesterday











                    3












                    $begingroup$

                    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









                    $endgroup$













                    • $begingroup$
                      No, the OP requested all six letter words.
                      $endgroup$
                      – m_goldberg
                      yesterday










                    • $begingroup$
                      Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                      $endgroup$
                      – Doorknob
                      yesterday
















                    3












                    $begingroup$

                    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









                    $endgroup$













                    • $begingroup$
                      No, the OP requested all six letter words.
                      $endgroup$
                      – m_goldberg
                      yesterday










                    • $begingroup$
                      Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                      $endgroup$
                      – Doorknob
                      yesterday














                    3












                    3








                    3





                    $begingroup$

                    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 yesterday









                    JagraJagra

                    7,88312159




                    7,88312159












                    • $begingroup$
                      No, the OP requested all six letter words.
                      $endgroup$
                      – m_goldberg
                      yesterday










                    • $begingroup$
                      Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                      $endgroup$
                      – Doorknob
                      yesterday


















                    • $begingroup$
                      No, the OP requested all six letter words.
                      $endgroup$
                      – m_goldberg
                      yesterday










                    • $begingroup$
                      Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                      $endgroup$
                      – Doorknob
                      yesterday
















                    $begingroup$
                    No, the OP requested all six letter words.
                    $endgroup$
                    – m_goldberg
                    yesterday




                    $begingroup$
                    No, the OP requested all six letter words.
                    $endgroup$
                    – m_goldberg
                    yesterday












                    $begingroup$
                    Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                    $endgroup$
                    – Doorknob
                    yesterday




                    $begingroup$
                    Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
                    $endgroup$
                    – Doorknob
                    yesterday


















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