6 boys and 6 girls sit alternately in x ways in a circular shape in y ways alternately then 
 

$$\begin{gathered} x=\mathrm{Number} \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{arranging} 6 \mathrm{boys} \mathrm{and} 6 \mathrm{girls} \mathrm{alternately} \mathrm{in} \mathrm{a} \mathrm{row} \\ =2{\left(6!\right)}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{y}=\mathrm{number} \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{arranging} 6 \mathrm{boys} \mathrm{and} 6 \mathrm{girls} \mathrm{around} \mathrm{a} \mathrm{circle} \mathrm{alternately} \\ =5!\left(6!\right) \end{gathered}$$

$\Rightarrow \frac{x}{y}=\frac{2\times 6!\times 6!}{6!\times 5!}=12\therefore \boxed{x=12y}$