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Four boys and Four girls are seated together in a row 03 Sep 2022, 03:50
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95% (hard)

Question Stats:

34% (02:18) correct 66% (02:03) wrong based on 29 sessions

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Four boys and Four girls are seated together in a row in such a way that exactly 2 out of the 4 boys are sitting together. In how many ways they can be seated?

A. 7! * 8
B. 6! * 18
C. 7! * 3
D. 6! * 24
E. 8! - (7! * 2)

(adapted from gmatfree)

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D
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SaquibHGMATWhiz
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Four boys and Four girls are seated together in a row 08 Sep 2022, 04:02
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    There are 4 boys and 4 girls. We need to make sure that exactly two boys are sitting together.

    So basically we are dealing with 7 entities B, B, BB, G, G, G, and G.

    We need to do it in a stepwise manner taking care of all the conditions.

      1. Since there is no issue with girls we can first arrange them in 4! ways and we have an arrangement such as _G_G_G_G_
      2. Now boys are grouped as BB, B, B so we need to first count in how many ways we can select 2 boys to form the entity BB. This can be done in 4C2 = 6 ways.
      3. Now since in BB, they are not identical so we need to arrange them as well and we can do that in 2! ways.
      4. Now we need to put these 3 entities of boys (BB, B, B) in 5 possible gaps. So using the filling space method we can do that in \(5 \times 4 \times 3 = 60\) ways


    The answer is \(4! \times 6 \times 2! \times 60\) = \(4! \times 720\) = \(6! \times 24\)

The answer is option D.
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Four boys and Four girls are seated together in a row 08 Sep 2022, 07:42
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Hello,

Can I ask what do you mean by the filling method? Thank you in advance
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Four boys and Four girls are seated together in a row 10 Sep 2022, 09:31
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I think it should be
7 ! X 4C2 X 2! ,
Nowhere it's given that two girls can't seat together

Posted from my mobile device
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Four boys and Four girls are seated together in a row 19 Sep 2022, 11:32
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SaquibHGMATWhiz
    There are 4 boys and 4 girls. We need to make sure that exactly two boys are sitting together.

    So basically we are dealing with 7 entities B, B, BB, G, G, G, and G.

    We need to do it in a stepwise manner taking care of all the conditions.

      1. Since there is no issue with girls we can first arrange them in 4! ways and we have an arrangement such as _G_G_G_G_
      2. Now boys are grouped as BB, B, B so we need to first count in how many ways we can select 2 boys to form the entity BB. This can be done in 4C2 = 6 ways.
      3. Now since in BB, they are not identical so we need to arrange them as well and we can do that in 2! ways.
      4. Now we need to put these 3 entities of boys (BB, B, B) in 5 possible gaps. So using the filling space method we can do that in \(5 \times 4 \times 3 = 60\) ways


    The answer is \(4! \times 6 \times 2! \times 60\) = \(4! \times 720\) = \(6! \times 24\)

The answer is option D.
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I am really struggling with this one... The condition(s), to me, are limited to just having 2 boys "grouped" as one. I am concluding that the answer would be something like 7! * 2.

could you (or anyone) elaborate on calculating the position of girls as 4! and putting the 3 entities in 5 gaps instead of 4?
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Four boys and Four girls are seated together in a row 19 Sep 2022, 13:10
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AryaSwagat
I think it should be
7 ! X 4C2 X 2! ,
Nowhere it's given that two girls can't seat together

Posted from my mobile device
Show more


You're correct that 2 girls can sit with each other. Three girls will also be sitting with each other in some of the arrangements.

There are 5 spots to place the block of two boys and each of the other two boys.

So after placing the the 3 into 5 spots, there will be 2 spots left unfilled.

These 2 spots being empty means that in some of the arrangements 2 or 3 girls will be sitting next to each other.

Only when a girl occupies each end will no girl be sitting next to another, because the boys will be sitting in the 3 spots between the 4 girls.
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Four boys and Four girls are seated together in a row 19 Sep 2022, 13:22
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wtstone
SaquibHGMATWhiz
    There are 4 boys and 4 girls. We need to make sure that exactly two boys are sitting together.

    So basically we are dealing with 7 entities B, B, BB, G, G, G, and G.

    We need to do it in a stepwise manner taking care of all the conditions.

      1. Since there is no issue with girls we can first arrange them in 4! ways and we have an arrangement such as _G_G_G_G_
      2. Now boys are grouped as BB, B, B so we need to first count in how many ways we can select 2 boys to form the entity BB. This can be done in 4C2 = 6 ways.
      3. Now since in BB, they are not identical so we need to arrange them as well and we can do that in 2! ways.
      4. Now we need to put these 3 entities of boys (BB, B, B) in 5 possible gaps. So using the filling space method we can do that in \(5 \times 4 \times 3 = 60\) ways


    The answer is \(4! \times 6 \times 2! \times 60\) = \(4! \times 720\) = \(6! \times 24\)

The answer is option D.

I am really struggling with this one... The condition(s), to me, are limited to just having 2 boys "grouped" as one. I am concluding that the answer would be something like 7! * 2.

could you (or anyone) elaborate on calculating the position of girls as 4! and putting the 3 entities in 5 gaps instead of 4?
Show more

Your 7! seems to come from the idea that there are 4 girls and 3 boy entities to be arranged, with the 2 boys contributing the 2 multiple.

But arranging this way allows for some arrangements where more than 2 boys sit next to each other.

The problem has the restriction that EXACTLY 2 boys sit next to each other. This means ONLY 2 boys can do so.

The boy entities need to be separated from each other by at least 1 girl.

The 3 boy entities can be placed at either end and between the girls.

There are 2 ends and 3 spaces between the girls, hence 5 available spots.

Posted from my mobile device

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