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 2 days ago









      J. M. is slightly pensive♦

      98.5k10308466




      98.5k10308466










      asked 2 days ago









      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
              2 days ago



















            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 2 days ago









                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
                      2 days ago
















                    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
                      2 days ago














                    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 2 days ago









                    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
                      2 days 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
                      2 days 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
                    2 days 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
                    2 days ago











                    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 2 days ago









                    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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Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029

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                    SlayerMusikgrupper bildade 1981Amerikanska thrash metal-grupper amerikansktthrash metalHuntington Park1981gitarristernaJeff HannemanKerry KingbasistenTom ArayatrummisenDave LombardoGary HoltPaul Bostaphcertifierats med guldMetallicaAnthraxMegadeththrash metal1986extrema metalgenrerdeath metal22 januari2018USAKanadaEuropaHuntington ParkKalifornien1981gitarristernaJeff HannemanKerry KingbassångDave Lombardoheavy metalbrittiskaIron MaidenJudas PriestthrashMetallicaRattSteeler1982demoundergroundMetal Blade1983high schoolGene HoglanDeathStrapping Young Ladkassettvinyl1984LondonMarquee Club1984Eldorado StudiosTrack Record StudiosLos Angelesmastrade1985CDCombat Recordsextreme metalVenomExodusStudio 54 club1985VHSCombat RecordsBelgienUFORick RubinDef Jam Recordship hoprockakterThe Black CrowesRun–D.M.C.LL Cool JBeastie BoysthrashbpmAngel of DeathRaining BloodAndy WallaceHowie Weinberg1986bildskivaBillboard 200RIAAMetal HammerKerrang!Angela Gossowdeath metalArch Enemythrash metalOverk...
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