Arrangements of all 26 letters [closed]












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$begingroup$


Consider all permutations of the twenty-six English letters that start with the letter $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?



I really couldn’t figure out how to do this one, can someone explain the solution






combinatorics permutations combinations







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closed as off-topic by user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg Feb 11 at 16:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    Consider all permutations of the twenty-six English letters that start with the letter $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?



    I really couldn’t figure out how to do this one, can someone explain the solution






    combinatorics permutations combinations







    share|cite|improve this question











    $endgroup$



    closed as off-topic by user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg Feb 11 at 16:46


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1


      0



      $begingroup$


      Consider all permutations of the twenty-six English letters that start with the letter $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?



      I really couldn’t figure out how to do this one, can someone explain the solution






      combinatorics permutations combinations







      share|cite|improve this question











      $endgroup$




      Consider all permutations of the twenty-six English letters that start with the letter $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?



      I really couldn’t figure out how to do this one, can someone explain the solution






      combinatorics permutations combinations




      combinatorics permutations combinations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 22:46









      GNUSupporter 8964民主女神 地下教會

      13.3k72549




      13.3k72549










      asked Jan 14 at 22:44









      user601297user601297

      37119




      37119




      closed as off-topic by user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg Feb 11 at 16:46


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg Feb 11 at 16:46


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Clearly $x$ must occur after $y$. So the options are





          • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;


          • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;

          • and so on;


          • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.


          Total, $23+21+cdots+1=144$. Then arrange the other $23$ letters. Answer
          $$144times23! .$$






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Clearly $x$ must occur after $y$. So the options are





            • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;


            • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;

            • and so on;


            • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.


            Total, $23+21+cdots+1=144$. Then arrange the other $23$ letters. Answer
            $$144times23! .$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Clearly $x$ must occur after $y$. So the options are





              • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;


              • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;

              • and so on;


              • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.


              Total, $23+21+cdots+1=144$. Then arrange the other $23$ letters. Answer
              $$144times23! .$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Clearly $x$ must occur after $y$. So the options are





                • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;


                • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;

                • and so on;


                • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.


                Total, $23+21+cdots+1=144$. Then arrange the other $23$ letters. Answer
                $$144times23! .$$






                share|cite|improve this answer









                $endgroup$



                Clearly $x$ must occur after $y$. So the options are





                • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;


                • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;

                • and so on;


                • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.


                Total, $23+21+cdots+1=144$. Then arrange the other $23$ letters. Answer
                $$144times23! .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 23:05









                DavidDavid

                68.9k667130




                68.9k667130















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