Number of ways in which they can be seated if the $2$ girls are together and the other two are also together...
$begingroup$
$5$ 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.
combinatorics
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
$endgroup$
add a comment |
$begingroup$
$5$ 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.
combinatorics
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$5$ 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.
combinatorics
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
$endgroup$
$5$ 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
combinatorics
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
edited Apr 4 '16 at 18:51
N. F. Taussig
44.1k93356
44.1k93356
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
asked Apr 4 '16 at 16:06
mathspuzzlemathspuzzle
25917
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $$15times 4!times 5!=43,200$$
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
$endgroup$
add a comment |
$begingroup$
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 $binom{6}{2}4!5!$.
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
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2 Answers
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2 Answers
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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 $$15times 4!times 5!=43,200$$
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
$endgroup$
add a comment |
$begingroup$
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 $$15times 4!times 5!=43,200$$
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
$endgroup$
add a comment |
$begingroup$
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 $$15times 4!times 5!=43,200$$
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
$endgroup$
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 $$15times 4!times 5!=43,200$$
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
edited Apr 4 '16 at 16:40
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
answered Apr 4 '16 at 16:18
lulululu
40.9k24879
40.9k24879
add a comment |
add a comment |
$begingroup$
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 $binom{6}{2}4!5!$.
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
$begingroup$
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 $binom{6}{2}4!5!$.
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
$begingroup$
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 $binom{6}{2}4!5!$.
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
$endgroup$
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 $binom{6}{2}4!5!$.
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
answered Apr 4 '16 at 16:35
André NicolasAndré Nicolas
452k36425810
452k36425810
add a comment |
add a comment |
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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