Group of two persons can be selected in ${\mathrm{}}^4{C}_2$ ways. Let $P_1{P}_2$ are together and $P_3{P}_4$ are together. Let $x_1$ be the number of seats vacant at the left end, $x_2$ seats are between two pairs and $x_3$ seats are vacant at the right end.

We have, $x_1+x_2+x_3=6,$ where $x_1, x_3\ge \mathrm{0 }\mathrm{and} x_2\ge 1$
$\Rightarrow x_1+\left(x_2+1\right)+x_3=6, \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e} x_1, x_2, x_3\ge 0$
$\Rightarrow x_1+x_2+x_3=5$
Number of ways of selecting seats $={\mathrm{}}^{n+r-1}{C}_{r-1}={\mathrm{}}^{3+5-1}{C}_{3-1}={\mathrm{}}^7{C}_2$
Persons can interchange their seats in $2!\mathrm{}\mathrm{}\times 2!$ ways
$\Rightarrow$ Required ways $={\mathrm{}}^4{C}_2\times {\mathrm{}}^7{C}_2\times 2!\mathrm{}\mathrm{}\times 2!={\mathrm{}}^4{P}_2\times {\mathrm{}}^7{P}_2$