In how many ways can 4 girls and 3 boys sit in a row such that just the girls are to sit next to each other? Answer: 288

Please explain how to get this.

I understand that we have

GGGG => 4 girls next to each other B B B => 3 boys

but how do you put them together and work out the number of possible ways. They are different so not identical

  • $\begingroup$ is it coincidence that $\ 4!*3! *2 = 288$ ? if you name each girl G1,G2,G3,G4 and try to rearrange them (ignoring the boys) there are $\ 4! \$ ways of arranging. $\endgroup$ – JLL Jun 15 '14 at 15:46
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    $\begingroup$ I'm getting $4!*4! = 576$. The girls are treated as a single unit. Each boy is a single unit by himself. Permute the four units = $4!$. Then permute the girls within the unit: $4!$. Multiply the two. I can't see how they get $288$, which is half the answer. $\endgroup$ – Deepak Jun 15 '14 at 15:51
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    $\begingroup$ in your question you mentioned there are 3 boys, not 4. $\endgroup$ – JLL Jun 15 '14 at 15:53
  • $\begingroup$ @JeanFerreira Yes 3 single-boy units + 1 unit of 4 girls = 4 units to be permuted. $\endgroup$ – Deepak Jun 15 '14 at 15:56
  • $\begingroup$ As stated, the solution given is half what the correct answer should be. $\endgroup$ – Namaste Jun 15 '14 at 16:00

We can have:


(There are four ways to place a group of four consecutive girls in a row of seven.

The girls can be permuted in each case $4!$, and so can the boys $(3!)$.

$$\bf 4\times 4!\times 3! = 576$$

NOTE If the intention of the author was that girls must sit next to a girl, and boys next to a boy, then there are only two ways to place the group of girls: GGGGBBB, BBBGGGG.

In that case, we have $2 \times 4!\times 3! = 288$.

  • $\begingroup$ I'm getting what you're getting, but the answer is supposed to be half of this. I think the given answer is wrong, or the conditions are imprecisely stated. $\endgroup$ – Deepak Jun 15 '14 at 15:53
  • $\begingroup$ I agree. As stated, this answer would be correct. If it was also intended that the boys must sit together, then it would be half that. $\endgroup$ – Namaste Jun 15 '14 at 15:56
  • $\begingroup$ Agreed but that stipulation wasn't stated in the question. The boys can sit anywhere in the group by default, in which case the answer is 576. $\endgroup$ – Deepak Jun 15 '14 at 16:00
  • $\begingroup$ I agree, @Deepak, which is why I posted this answer! ;-) $\endgroup$ – Namaste Jun 15 '14 at 16:02
  • $\begingroup$ Thanks. This is exactly what I did and this is why I posted the question. It's not the first time my stats lecturer has put the incorrect answer on a tut question. $\endgroup$ – StephanCasey Jun 15 '14 at 16:05

To be honest I am not sure if this is the right way to do it, some one please correct me if Im wrong.

By thinking about the group of girls and boys separately, (as mentioned in the comment):

girls = $\ G_1,G_2,G_3,G_4$; boys = $\ B_1,B_2,B_3$

we can create $\ 4$! ways of arranging the girls(by themselves) and 3! possibilities of arranging the boys by themselves.

now if we put them together, we are doubling the possibilites of arrangement.

thus, # of arrangements = $\ 4!*3! *2 =288 $

  • $\begingroup$ You counted the number of arrangements in which all the girls sit together and all the boys sit together. If that were the problem, the factor of $2$ would represent whether the boys sat to the left or to the right of the girls. $\endgroup$ – N. F. Taussig Apr 5 '16 at 16:22

I can explain why this answer is 288 and not 576. The difference between it being 288 and 576 is based on one word in the question: "just"

If the boys and girls are arranged where just the girls (and not the boys) are sitting next to each other, then the problem would be: 2 x 4! x 3! which = 288 instead of 4 x 4! x 3! which is 576.

Why 2 instead of 4? If you have three boys 1 2 3, then there's 2 spaces the girls could fit within them (between the 1 and 2, and between the 2 and 3). If the girls are placed before the 1 or after the 3, then it allows the boys to automatically be together which the question says just the girls should sit next to each other.

I hope that explains it. My text book had 2 variations of the same question where one was "In how many ways can they sit in a row if the girls are to sit together?" and "In how many ways can they sit in a row if just the girls are to sit together. Both questions had different answers like the ones above.


I doubt whether you have worded your question correctly as according to your question the answer should come out as 576 and not 288. Anyway an approach that could lead to 288 is as follows:

Let's number girls and boys as : G1, G2, G3, G4 and B1, B2, B3 respectively.

Since a girl has to sit next to a girl only therefore all the girls would sit together G1,G2,G3,G4 or G2,G3,G1,G4 etc. Now these girls can be arranged in 4! = 24 ways.

Now we have 3 boys who can be made to sit together in 3! = 6 ways.

Finally we have Girls (G) and Boys (B) who have to be seated together and they can be seated in 2! = 2 ways (consider 4 girls as one group = G and 3 boys as one group = B)

From above three steps we get the total number of ways as : 4! * 3! * 2! = 288

  • $\begingroup$ Your method misses out the possibilities where the group of girls can be sandwiched between the boys, e.g. one boy on the left, then the four girls, then two boys on the right. I think this is allowed by the stipulation of the question. $\endgroup$ – Deepak Jun 15 '14 at 15:55
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    $\begingroup$ "Just the girls are to sit together." There is no requirement regarding the placement of the boys. $\endgroup$ – Namaste Jun 15 '14 at 15:55
  • $\begingroup$ Deepak, you keep changing your question, read it carefully. $\endgroup$ – JLL Jun 15 '14 at 15:56
  • $\begingroup$ @JeanFerreira I have read it carefully. How am I changing the question? $\endgroup$ – Deepak Jun 15 '14 at 15:59
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    $\begingroup$ @JeanFerreira No worries. Yes, it will double the answer to $576$ as amWhy and I have shown. We think this should be the correct answer. Sometimes (often?!) book answers are not correct. $\endgroup$ – Deepak Jun 15 '14 at 16:04

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