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





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "387"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193857%2fcreate-all-possible-words-using-a-set-or-letters%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematica Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193857%2fcreate-all-possible-words-using-a-set-or-letters%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

                    Image
                    Engineer refusing to file/disclose patents Extending the spectral theorem for bounded self adjoint operators to bounded normal operators Do varchar(max), nvarchar(max) and varbinary(max) columns affect select queries? Why is it that I can sometimes guess the next note? What linear sensor for a keyboard? Why don’t women say שלא עשני גויה and שפחה, according to Ashkenazim? Did arcade monitors have same pixel aspect ratio as TV sets? anything or something to eat Query about absorption line spectra Can I use my Chinese passport to enter China after I acquired another citizenship? Fuse symbol on toroidal transformer Create all possible words using a set or letters Does the Mind Blank spell prevent the target from being frightened? Can I sign legal documents with a smiley face? When quoting, must I also copy hyphens used to divide words that continue on the next line? Hot bath for aluminium engine block and heads Journal losing indexing services Is it possib
                    Read more

                    Bunad

                    Image
                    Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
                    Read more

                    Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

                    Image
                    FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mörg þeirra voru seld til Bretlands til að vinna í kolanámum og um 1960 voru aðeins nokkrir hestar eftir. Átak var gert í að varðveita hestinn og eru nú til 70-80 hross í Færeyjum. [1] Heimild |   Wikimedia Commons er með margmiðlunarefni sem te
                    Read more