Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

$\text{ Let }{\mathrm{A}}_1,{ \mathrm{A}}_2,{ \mathrm{A}}_3\text{ be members of one family }$

$\text{ Let }{\mathrm{B}}_1,{ \mathrm{B}}_2,{ \mathrm{B}}_3\text{ be members of another family }$

$\text{ Let }{\mathrm{C}}_1,{\mathrm{C}}_2,{\mathrm{C}}_3,{\mathrm{C}}_4\text{ be members of another family }$

$\begin{array}{l}\text{ Number of ways of arranging }3\text{ units }=\left(3{\mathrm{P}}_3\right)=\overline{)3}\\ \text{ Number of internal arrangements of }{\mathrm{A}}_1,{ \mathrm{A}}_2,{ \mathrm{A}}_3=\left(3{\mathrm{P}}_3\right)=\overline{)3}\\ \text{ Number of internal arrangements of }{\mathrm{B}}_1,{ \mathrm{B}}_2,{ \mathrm{B}}_3=\left(3{\mathrm{P}}_3\right)=\overline{)3}\\ \text{ Number of internal arrangements of }{\mathrm{C}}_1,{\mathrm{C}}_2,{\mathrm{C}}_3,{\mathrm{C}}_4=\left(4{\mathrm{P}}_4\right)=\overline{)4}\end{array}$

 Total Number of arrangements, so that members of same family are not separated

$=\left(\overline{)3}\times \overline{)3}\times \overline{)3}\times \overline{)4}\right)={\left(\overline{)3}\right)}^3.\overline{)4}$