Assumptions:
- Let’s say that the seats are numbered $1,2,\dots,8$.
- Let’s say the boys will sit in the odd-numbered seats: $1,3,5,7$. Girls will sit in the even-numbered seats: $2,4,6,8$.
For seat #$1$: you can choose $1$ boy out of $4$ boys in $4$ ways.
For seat #$3$: you can choose $1$ boy out of $3$ remaining boys in $3$ ways.
For seat #$5$: you can choose $1$ boy out of $2$ remaining boys in $2$ ways.
For seat #$7$: you can choose $1$ boy out of $1$ remaining boy in $1$ way.
The number of ways you can seat the boys is $4! = 4\cdot 3\cdot 2\cdot 1\tag{a}$
You can do the same for girls to obtain:
The number of ways you can seat the girls is $4! = 4\cdot 3\cdot 2\cdot 1\tag{b}$
So, the number of distinct ways to seat the boys AND the girls is the product of $(a)$ and $(b)$:
$4! \cdot 4!\tag{c}$
Now, remember our assumptions 1 and 2 above? These assumptions cover only $1$ of $2$ ways we can run the experiment. That is, there is another way in which girls sit in the odd-numbered chairs and boys sit in the even-numbered chairs. The number of ways resulting from this new assumption is exactly the same as the one found in $(c)$ above.
So the total number of ways $=4! \cdot 4! + 4! \cdot 4! = 2 \cdot 4! \cdot 4!\;.$
Edit: As per Brian M. Scott comment below, this solution is incomplete.
We need to subtract the number of ways the given pair can sit together from the number shown above.