If you arrange as GGBBBBBG, there will be $5$ boys sitting between $2$ girls, but the question demands "no boys between 2 girls"
You are not puzzled by the sum, it's the language of the question which is driving you to wrong way. Don't be upset. I also get puzzled. Recently, in an exam I didn't attend an easy probability sum for its puzzling words and did another sum which was a page long, lol.
You can do it as mentioned below:
First make all 5 boys seated, which can be done in $5!$ ways. Now you can see 6 gaps created.......(1)
$$\underline{} B \underline{} B \underline{} B \underline{} B \underline{} B \underline{}$$
Now you take group of girls together. Consider GGG. You can make this group seat in any one of the $6$ gaps, so there are $6$ ways and you can arrange the $3$ girls among themselves in $3!$ ways , so here total no. of ways $6 \cdot 3!$
........................(2)
From (1) and (2) you can say no. of ways $= 5! \cdot 6 \cdot 3!$
If there's no restriction, then total no. of ways $=8!$.
Divide to get the probability
$$\frac{5! \cdot 6 \cdot 3!}{8!} = \frac{6! \cdot 3!}{8!} = \frac{6}{8 \cdot 7} = \frac{3}{28}$$
!Answer!
Best Answer
Imagine John being part of the committee. You might have selected him out of the seven men at first, and then have put three other people with him afterward (let's call them Adam, Mark and Sarah). Alternatively, it is possible that you selected Adam first, but put John in as part of your second selection process (which also included Mark and Sarah). Because both committees are the same, you have indeed overcounted. Note that you are not overcounting in the alternative approach, because each person is selected (or not) in one single step.