Number of ways in which they can be seated if the $2$ girls are together and the other two are also together...

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$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.





I divided the $4$ girls in two groups in $frac{4!}{2!2!}$ ways.


I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=frac{4!}{2!2!}times 10times 5!=7200$


But the answer in the book is $43200$. I don't know where I am wrong.










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  • $begingroup$
    What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:10










  • $begingroup$
    Sorry,"the" was not given,i edited it.@lulu
    $endgroup$
    – mathspuzzle
    Apr 4 '16 at 16:13










  • $begingroup$
    no problem. I'll post something below.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:13
















1












$begingroup$


$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.





I divided the $4$ girls in two groups in $frac{4!}{2!2!}$ ways.


I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=frac{4!}{2!2!}times 10times 5!=7200$


But the answer in the book is $43200$. I don't know where I am wrong.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:10










  • $begingroup$
    Sorry,"the" was not given,i edited it.@lulu
    $endgroup$
    – mathspuzzle
    Apr 4 '16 at 16:13










  • $begingroup$
    no problem. I'll post something below.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:13














1












1








1


2



$begingroup$


$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.





I divided the $4$ girls in two groups in $frac{4!}{2!2!}$ ways.


I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=frac{4!}{2!2!}times 10times 5!=7200$


But the answer in the book is $43200$. I don't know where I am wrong.










share|cite|improve this question











$endgroup$




$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.





I divided the $4$ girls in two groups in $frac{4!}{2!2!}$ ways.


I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=frac{4!}{2!2!}times 10times 5!=7200$


But the answer in the book is $43200$. I don't know where I am wrong.







combinatorics






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edited Apr 4 '16 at 18:51









N. F. Taussig

44.1k93356




44.1k93356










asked Apr 4 '16 at 16:06









mathspuzzlemathspuzzle

25917




25917












  • $begingroup$
    What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:10










  • $begingroup$
    Sorry,"the" was not given,i edited it.@lulu
    $endgroup$
    – mathspuzzle
    Apr 4 '16 at 16:13










  • $begingroup$
    no problem. I'll post something below.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:13


















  • $begingroup$
    What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:10










  • $begingroup$
    Sorry,"the" was not given,i edited it.@lulu
    $endgroup$
    – mathspuzzle
    Apr 4 '16 at 16:13










  • $begingroup$
    no problem. I'll post something below.
    $endgroup$
    – lulu
    Apr 4 '16 at 16:13
















$begingroup$
What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
$endgroup$
– lulu
Apr 4 '16 at 16:10




$begingroup$
What do you mean "the two girls are together"? That makes it sound as if the pair is specified.
$endgroup$
– lulu
Apr 4 '16 at 16:10












$begingroup$
Sorry,"the" was not given,i edited it.@lulu
$endgroup$
– mathspuzzle
Apr 4 '16 at 16:13




$begingroup$
Sorry,"the" was not given,i edited it.@lulu
$endgroup$
– mathspuzzle
Apr 4 '16 at 16:13












$begingroup$
no problem. I'll post something below.
$endgroup$
– lulu
Apr 4 '16 at 16:13




$begingroup$
no problem. I'll post something below.
$endgroup$
– lulu
Apr 4 '16 at 16:13










2 Answers
2






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2












$begingroup$

Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2times 6!=3600$$ suitable arrangements.



Now, sticking to these pairs but varying the order gets us $$4times 3600=14,400$$



And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3times 14,400= 43,200$$



Method II: (closer to what you were trying) Arrange the kids as $$-;AB-CD;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $${0,5,0},;{1,4,0},;{0,4,1},;{2,3,0},;{0,3,2},;{1,3,1},;{3,2,0},;{0,2,3},;{2,2,1},;{1,2,2},;{3,1,1},;{1,1,3},{2,1,2},;{4,1,0},;{0,1,4}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15times 4!times 5!=43,200$$






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$endgroup$





















    6












    $begingroup$

    First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them.
    $$text{b}qquadtext{b}qquadtext{b}qquadtext{b}qquadtext{b}$$
    These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $binom{6}{2}$ ways.



    We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $binom{6}{2}4!5!$.






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      2 Answers
      2






      active

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      2 Answers
      2






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      $begingroup$

      Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2times 6!=3600$$ suitable arrangements.



      Now, sticking to these pairs but varying the order gets us $$4times 3600=14,400$$



      And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3times 14,400= 43,200$$



      Method II: (closer to what you were trying) Arrange the kids as $$-;AB-CD;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $${0,5,0},;{1,4,0},;{0,4,1},;{2,3,0},;{0,3,2},;{1,3,1},;{3,2,0},;{0,2,3},;{2,2,1},;{1,2,2},;{3,1,1},;{1,1,3},{2,1,2},;{4,1,0},;{0,1,4}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15times 4!times 5!=43,200$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2times 6!=3600$$ suitable arrangements.



        Now, sticking to these pairs but varying the order gets us $$4times 3600=14,400$$



        And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3times 14,400= 43,200$$



        Method II: (closer to what you were trying) Arrange the kids as $$-;AB-CD;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $${0,5,0},;{1,4,0},;{0,4,1},;{2,3,0},;{0,3,2},;{1,3,1},;{3,2,0},;{0,2,3},;{2,2,1},;{1,2,2},;{3,1,1},;{1,1,3},{2,1,2},;{4,1,0},;{0,1,4}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15times 4!times 5!=43,200$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2times 6!=3600$$ suitable arrangements.



          Now, sticking to these pairs but varying the order gets us $$4times 3600=14,400$$



          And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3times 14,400= 43,200$$



          Method II: (closer to what you were trying) Arrange the kids as $$-;AB-CD;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $${0,5,0},;{1,4,0},;{0,4,1},;{2,3,0},;{0,3,2},;{1,3,1},;{3,2,0},;{0,2,3},;{2,2,1},;{1,2,2},;{3,1,1},;{1,1,3},{2,1,2},;{4,1,0},;{0,1,4}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15times 4!times 5!=43,200$$






          share|cite|improve this answer











          $endgroup$



          Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2times 6!=3600$$ suitable arrangements.



          Now, sticking to these pairs but varying the order gets us $$4times 3600=14,400$$



          And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3times 14,400= 43,200$$



          Method II: (closer to what you were trying) Arrange the kids as $$-;AB-CD;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $${0,5,0},;{1,4,0},;{0,4,1},;{2,3,0},;{0,3,2},;{1,3,1},;{3,2,0},;{0,2,3},;{2,2,1},;{1,2,2},;{3,1,1},;{1,1,3},{2,1,2},;{4,1,0},;{0,1,4}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15times 4!times 5!=43,200$$







          share|cite|improve this answer














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          edited Apr 4 '16 at 16:40

























          answered Apr 4 '16 at 16:18









          lulululu

          40.9k24879




          40.9k24879























              6












              $begingroup$

              First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them.
              $$text{b}qquadtext{b}qquadtext{b}qquadtext{b}qquadtext{b}$$
              These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $binom{6}{2}$ ways.



              We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $binom{6}{2}4!5!$.






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them.
                $$text{b}qquadtext{b}qquadtext{b}qquadtext{b}qquadtext{b}$$
                These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $binom{6}{2}$ ways.



                We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $binom{6}{2}4!5!$.






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them.
                  $$text{b}qquadtext{b}qquadtext{b}qquadtext{b}qquadtext{b}$$
                  These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $binom{6}{2}$ ways.



                  We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $binom{6}{2}4!5!$.






                  share|cite|improve this answer









                  $endgroup$



                  First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them.
                  $$text{b}qquadtext{b}qquadtext{b}qquadtext{b}qquadtext{b}$$
                  These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $binom{6}{2}$ ways.



                  We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $binom{6}{2}4!5!$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 4 '16 at 16:35









                  André NicolasAndré Nicolas

                  452k36425810




                  452k36425810






























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