A family consisting of a mother, father and their 8 children ( 4 boys and 4 girls) are to be seated at a round table in a party. How many ways can this be done if the mother and father sit together and the males and females alternate?

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\overrightarrow{\mathrm{a}}|=\sqrt{31}, 4|\overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{c}}|=2$ and $2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})=3(\overline{\mathrm{c}} \times \overline{\mathrm{a}})$ and if the angle between $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is $\frac{2 \pi}{3}$ then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$

Let $\bar{a}=2 \hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-\hat{k}$ and vector $\bar{c}$ be coplanar. If $\bar{c}$ is perpendicular to $\bar{a}$, then $\bar{c}$ is

If $x=-2+\sqrt{-3}$, then the value of $2 x^4+5 x^3+7 x^2-x+38$ is equal to