Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never...












3












$begingroup$



Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.




Attempt:



total number of ways - number of ways in which all girls are together



$= 12! - 2!times(6!)times (6!)$



But answer given is $12! - 7!6!$



Is the given answer wrong?






combinatorics







share|cite|improve this question









$endgroup$

















    3












    $begingroup$



    Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.




    Attempt:



    total number of ways - number of ways in which all girls are together



    $= 12! - 2!times(6!)times (6!)$



    But answer given is $12! - 7!6!$



    Is the given answer wrong?






    combinatorics







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$



      Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.




      Attempt:



      total number of ways - number of ways in which all girls are together



      $= 12! - 2!times(6!)times (6!)$



      But answer given is $12! - 7!6!$



      Is the given answer wrong?






      combinatorics







      share|cite|improve this question









      $endgroup$





      Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.




      Attempt:



      total number of ways - number of ways in which all girls are together



      $= 12! - 2!times(6!)times (6!)$



      But answer given is $12! - 7!6!$



      Is the given answer wrong?






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 25 '18 at 12:04









      AbcdAbcd

      3,05531237




      3,05531237






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          With any constraint the number of possible combination is $12!$



          If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways



          and again the $6$ girls can be arranged in $6!$ ways






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Reminds me of math.stackexchange.com/questions/697433/…
            $endgroup$
            – lab bhattacharjee
            Oct 25 '18 at 12:47



















          2












          $begingroup$

          All the girls together means a sequence of 6 girls in a row.
          This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.



          There are $6!$ ways to place the girls and $6!$ ways to place the boys.



          So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.



          Thus there are $12! - 7! cdot 6!$ ways that respect the rule.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            With any constraint the number of possible combination is $12!$



            If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways



            and again the $6$ girls can be arranged in $6!$ ways






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Reminds me of math.stackexchange.com/questions/697433/…
              $endgroup$
              – lab bhattacharjee
              Oct 25 '18 at 12:47
















            4












            $begingroup$

            With any constraint the number of possible combination is $12!$



            If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways



            and again the $6$ girls can be arranged in $6!$ ways






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Reminds me of math.stackexchange.com/questions/697433/…
              $endgroup$
              – lab bhattacharjee
              Oct 25 '18 at 12:47














            4












            4








            4





            $begingroup$

            With any constraint the number of possible combination is $12!$



            If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways



            and again the $6$ girls can be arranged in $6!$ ways






            share|cite|improve this answer









            $endgroup$



            With any constraint the number of possible combination is $12!$



            If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways



            and again the $6$ girls can be arranged in $6!$ ways







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 25 '18 at 12:09









            lab bhattacharjeelab bhattacharjee

            226k15157275




            226k15157275












            • $begingroup$
              Reminds me of math.stackexchange.com/questions/697433/…
              $endgroup$
              – lab bhattacharjee
              Oct 25 '18 at 12:47


















            • $begingroup$
              Reminds me of math.stackexchange.com/questions/697433/…
              $endgroup$
              – lab bhattacharjee
              Oct 25 '18 at 12:47
















            $begingroup$
            Reminds me of math.stackexchange.com/questions/697433/…
            $endgroup$
            – lab bhattacharjee
            Oct 25 '18 at 12:47




            $begingroup$
            Reminds me of math.stackexchange.com/questions/697433/…
            $endgroup$
            – lab bhattacharjee
            Oct 25 '18 at 12:47











            2












            $begingroup$

            All the girls together means a sequence of 6 girls in a row.
            This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.



            There are $6!$ ways to place the girls and $6!$ ways to place the boys.



            So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.



            Thus there are $12! - 7! cdot 6!$ ways that respect the rule.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              All the girls together means a sequence of 6 girls in a row.
              This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.



              There are $6!$ ways to place the girls and $6!$ ways to place the boys.



              So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.



              Thus there are $12! - 7! cdot 6!$ ways that respect the rule.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                All the girls together means a sequence of 6 girls in a row.
                This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.



                There are $6!$ ways to place the girls and $6!$ ways to place the boys.



                So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.



                Thus there are $12! - 7! cdot 6!$ ways that respect the rule.






                share|cite|improve this answer









                $endgroup$



                All the girls together means a sequence of 6 girls in a row.
                This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.



                There are $6!$ ways to place the girls and $6!$ ways to place the boys.



                So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.



                Thus there are $12! - 7! cdot 6!$ ways that respect the rule.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 25 '18 at 12:10









                RonaldRonald

                673510




                673510






























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