Circular table problem with condition

I just started my combinatorics class and want to check if my reasoning is correct

there are 8 boys and 4 girls. How many ways can they be seated at a round table if there must always be two boys between any two consecutive girls?

My reasoning is we treat the $8$ boys as pairs, so we have $4$ pairs of boys to place say $x_1,x_2,x_3,x_4$, and $y_1,y_2,y_3,y_4$ girls. Since it is a round table there are $\dfrac{8!}{8}$ ways to do this. Then from here, I am not sure. I want to say the pair of boys can rearrange themselves 4! ways and similar for the girls hence the final answer should be $\dfrac{8!}{8}*(4!)^2?$

You can name the boys $A,B,\ldots, H$ and seat them around the table in $\frac{8!}{8}$ ways. Then girls can be seated between $(B,C)$, $(D,E)$, $(F,G)$, $(H,A)$. This can be done in $4!$ ways. Another possibility is, girls are seated between $(A,B)$, $(C,D)$, $(E,F)$, $(G,H)$. This can be achieved in $4!$ ways. Hence total is $$\frac{8!}{8} \times (4!+4!)$$

Alternatively we can seat the girls first. Recall the $12$ positions of numbers of a clock. We let the girls seat at positions $12,3,9,6$. There are $\frac{4!}{4}$ ways. The boys occupy the positions of remaining $8$ numbers in $8!$ ways. So answer is $$\frac{4!}{4}\times 8! = 3!\times 8!$$ same as above.