In how many ways 2 girls and 3 Boys can be seated in two cars heach having numbered seats 1 in the front and 3 at the back?how many seating arrangements are possible if 2 girls sit together in a back row on adjacent seats?

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$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{We} \mathrm{have}: \\ 2 \mathrm{girls}+3 \mathrm{boys}=5 \mathrm{people} \\ \mathrm{Total} \mathrm{seats} \mathrm{in} \mathrm{one} \mathrm{car}=1+3=4 \\ \mathrm{Total} \mathrm{seats} \mathrm{in} 2 \mathrm{cars}=4+4=8 \\ \mathrm{So} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{arrange} 5 \mathrm{people} \mathrm{at} 8 \mathrm{places}. \\ \Rightarrow \mathrm{Number} \mathrm{of} \mathrm{ways}{=}^8{\mathrm{P}}_5=\frac{8!}{\left(8-5\right)!} \\ =\frac{8!}{3!} \\ =\frac{8.7.6.5.4.3!}{3!} \\ =8.7.6.5.4 \\ =\mathbf{6720}\mathbf{ }\left(\mathbf{Answer}\right) \\ \left(\mathrm{ii}\right) \\ \mathrm{Now} \mathrm{first} \mathrm{of} \mathrm{all} 2 \mathrm{girls} \mathrm{will} \mathrm{have} \mathbf{2}\mathbf{ }\mathbf{options} \mathrm{i}.\mathrm{e}. \mathrm{either} \mathrm{to} \mathrm{choose} \mathrm{car} 1 \mathrm{or} \mathrm{car} 2. \\ \mathrm{Now} \mathrm{there} \mathrm{are} \mathrm{three} \mathrm{seats} \mathrm{in} \mathrm{back} \mathrm{row} \mathrm{but} \mathrm{as} \mathrm{we} \mathrm{want} \mathrm{them} \mathrm{to} \mathrm{sit} \mathrm{adjacent} \mathrm{to} \\ \mathrm{each} \mathrm{other}, \mathrm{hence} \mathrm{will} \mathrm{have} \mathbf{2}\mathbf{ }\mathbf{options} \mathrm{i}.\mathrm{e} \mathrm{Either} \mathrm{choose} \mathrm{seat} 1 \mathrm{and} 2 \mathrm{or} \mathrm{seat} 2 \mathrm{and} 3. \\ \mathrm{Now} \mathrm{there} \mathrm{will} \mathrm{be} \mathbf{2}\mathbf{!}\mathbf{ }\mathbf{arrangements} \mathrm{of} 2 \mathrm{girls}. \\ \mathrm{Now} \mathrm{we} \mathrm{are} \mathrm{left} \mathrm{to} \mathrm{arrange} \mathbf{3}\mathbf{ }\mathbf{people}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{6}\mathbf{ }\mathbf{places} \left(\mathrm{As} 2 \mathrm{places} \mathrm{are} \mathrm{occupied} \mathrm{by} \mathrm{girls}\right) \\ \Rightarrow \mathrm{Number} \mathrm{of} \mathrm{ways}=2\times 2\times 2!{\times }^6{\mathrm{P}}_3 \\ =8.\frac{6!}{\left(6-3\right)!} \\ =8.\frac{6!}{3!} \\ =8.\frac{6.5.4.3!}{3!} \\ =8.6.5.4 \\ =\mathbf{960}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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