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The question requires the lineup to be either BBBBGGGG or GGGGBBBB. I'm not sure if my thinking is correct, but this is my work so far:

Boys (notated as $B_1$ to $B_4$):

The first boy has 8 spots to stand in, therefore $P(B_1)$= $\frac 18$.

The second boy has 3 spots to stand in, therefore $P(B_2)$= $\frac 13$.

The third boy has 2 spots to stand in, therefore $P(B_3)$= $\frac 12$.

The fourth boy has only 1 spot to stand in, therefore $P(B_4)$= 1.

Girls (notated as $G_1$ to $G_4$):

The first girl only has 4 spots to stand in since the 4 boys occupied 4 spots already, therefore $P(G_1)$= $\frac 14$.

The second girl then has 3 spots to stand in, therefore $P(G_2)$= $\frac 13$.

The third girl has 2 spots to stand in, therefore $P(G_3)$= $\frac 12$.

The fourth girl has only 1 spot to stand in, therefore $P(G_4)$= 1.

Then $P(BBBBGGGG)$= ($\frac 18$ $\cdot$ $\frac 13$ $\cdot$$\frac 12$ $\cdot$ $1$ $\cdot$ $\frac 14$ $\cdot$ $\frac 13$ $\cdot$ $\frac 12$ $\cdot$ $1$) = $.000868$, which is the same for $P(GGGGBBBB)$, therefore the final answer is $0.001736$.

Is this correct or did I misunderstand anything?

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    $\begingroup$ Not sure of your notation. What is $P(B_1)$? In any case, any arrangement of the kids is determined by the locations of the girls...tho place the girls you need to pick four seats out of eight, so there are $\binom 84=70$ arrangements. Only two work for you, so the answer is $\frac 2{70}\approx 0.028571429$. $\endgroup$ – lulu Sep 29 '18 at 18:52
  • $\begingroup$ @lulu By $\frac{2}{70}$, do you mean that only 2 arrangements will give all 4 girls together? But how about the boys? $\endgroup$ – peco Sep 29 '18 at 19:01
  • $\begingroup$ No...I meant what you said. There are exactly two arrangements in which all the boys AND all the girls are together. $\endgroup$ – lulu Sep 29 '18 at 19:04
  • $\begingroup$ I think I understand now, thanks! $\endgroup$ – peco Sep 29 '18 at 19:05
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There are $2$ cases: $B_1B_2B_3B_4G_1G_2G_3G_4$ or $G_1G_2G_3G_4B_1B_2B_3B_4$,and the probability for each is $\dfrac{4!\times 4!}{8!}$. Thus the probability boys together and girls together is: $\dfrac{2\times 4!\times4!}{8!}$

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  • $\begingroup$ Just to confirm, having $\frac{4!}{8!}$ will be the probability of all 4 boys, for example, together? $\endgroup$ – peco Sep 29 '18 at 19:06
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    $\begingroup$ @peco: no you have to account for all positions. You also have to account for the girls' order, even when do not stand together. So it is actually $5 \times \frac{4!4!}{8!}$ $\endgroup$ – Alex Sep 29 '18 at 20:00

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