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Create all possible words using a set or letters

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Create all possible words using a set or letters


Finding all length-n words on an alphabet that have a specified number of each letterFinding all dictionary words that can be made with a given set of characters (Wordfeud/Scrabble)How to enumerate all possible binary associations?Sorting an Array with words in different languagesUsing StringCases and treating certain phrases as single wordsGraph showing valid English words obtained by insertion of single charactersTrim a list of elementsList all possible microstates and corresponding energy using mathematica.Selecting words having a specific number of letters from a textSelecting elements using two lists













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 2 hours 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









          $endgroup$













          • $begingroup$
            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
            7 mins ago



















          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|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









              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









              $endgroup$













              • $begingroup$
                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
                7 mins ago
















              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









              $endgroup$













              • $begingroup$
                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
                7 mins ago














              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












              • $begingroup$
                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
                7 mins 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
                7 mins 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
              7 mins 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
              7 mins ago











              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|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 10 mins ago









                          JagraJagra

                          7,85312159




                          7,85312159






























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