So I tried this question but somehow I have a hard time understanding what they ask .
The question goes : calculate the number of ways 3 girls and 4 boys can be seated in a row of seven(7) chairs if the arrangement is symmetrical.
My attempt I am unsure what they mean by symmetrical but i can only think of this :
B G B G B G B as an arrangement And then 4 . 2 . 3 . 3 . 2 .1 .1 =144
Note that the arrangements are $$\mbox{BBGGGBB}\\\mbox{BGBGBGB}\\\mbox{GBBGBBG}$$The centre chair cannot be occupied by a boy as then there will be 3 boys left who cannot be placed symmetrically on both sides of the central chair.
So the girl for the central chair can be selected in $3$ ways.
Now arrange $1$ girl and $2$ boys on each side of the central chair. They can be arranged in $\dbinom{2}{1}\times\dbinom{4}{2}=12$ ways
The remaining will go to the right side. They can be arranged in $3!=6$ ways
On the right the relative position of girls and boys will be same as on left and the girls and the boys can be arranged among themselves in $1!\times2! = 2$ ways.
Therefore, the total number of arrangements $=3\times12\times6\times2=432$
Good start! However, there are two other symmetrical arrangements: GBBGBBG and BBGGGBB. Each of these have 4!3! = 144 permutations as well, so your final answer should be 144*3 = 432.
