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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.










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.










    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.










      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago









      J. M. is slightly pensive

      98.3k10307466




      98.3k10307466










      asked 3 hours ago









      mf67mf67

      976




      976




















          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
            1 hour 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
                1 hour 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
                1 hour 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 3 hours 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
                1 hour 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
                1 hour 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
              1 hour 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
              1 hour 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 3 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 1 hour ago









                          JagraJagra

                          7,85312159




                          7,85312159



























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