Create all possible words using a set or letters












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 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





              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









              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






























                              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

                              Knooppunt Holsloot

                              Altaar (religie)

                              Gregoriusmis