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.

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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

























    up vote
    2
    down vote

    favorite













    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























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      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














      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 1 hour ago









      Abcd

      2,75011129




      2,75011129




















          2 Answers
          2






          active

          oldest

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          up vote
          3
          down vote



          accepted










          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




















          • Reminds me of math.stackexchange.com/questions/697433/…
            – lab bhattacharjee
            50 mins ago

















          up vote
          2
          down vote













          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




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            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




















            • Reminds me of math.stackexchange.com/questions/697433/…
              – lab bhattacharjee
              50 mins ago














            up vote
            3
            down vote



            accepted










            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




















            • Reminds me of math.stackexchange.com/questions/697433/…
              – lab bhattacharjee
              50 mins ago












            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            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












            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 1 hour ago









            lab bhattacharjee

            218k14153268




            218k14153268











            • Reminds me of math.stackexchange.com/questions/697433/…
              – lab bhattacharjee
              50 mins ago
















            • Reminds me of math.stackexchange.com/questions/697433/…
              – lab bhattacharjee
              50 mins ago















            Reminds me of math.stackexchange.com/questions/697433/…
            – lab bhattacharjee
            50 mins ago




            Reminds me of math.stackexchange.com/questions/697433/…
            – lab bhattacharjee
            50 mins ago










            up vote
            2
            down vote













            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
























              up vote
              2
              down vote













              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






















                up vote
                2
                down vote










                up vote
                2
                down vote









                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












                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 1 hour ago









                Ronald

                643510




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