A double decker bus has 15 seats in the lower deck and 13 seats in the upper deck. In how many ways can a marriage party of 28 persons be arranged if 4 old people refuse to go to the upper deck and 4 children wish to travel in the upper deck only.

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{seats} \mathrm{in} \mathrm{lower} \mathrm{deck}=15. \\ \mathrm{since} \mathrm{four} \mathrm{old} \mathrm{people} \mathrm{refuse} \mathrm{to} \mathrm{sit} \mathrm{in} \mathrm{upper} \mathrm{deck} \\ \mathrm{Remaining} \mathrm{seats} \mathrm{in} \mathrm{lower} \mathrm{deck} =15-4=11 and \\ \mathrm{Number} \mathrm{of} \mathrm{seats} \mathrm{in} \mathrm{upper} \mathrm{deck}=13. \\ Since 4 \mathrm{children} \mathrm{wish} \mathrm{to} \mathrm{sit} \mathrm{in} \mathrm{upper} \mathrm{deck} \\ \mathrm{Remaining} \mathrm{seats} \mathrm{in} \mathrm{upper} \mathrm{deck} =13-4=9 \\ \mathrm{No} \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{arranging} \mathrm{the} \mathrm{passengers} \\ =\mathrm{dividing} 20 \mathrm{people} \mathrm{into} \mathrm{two} \mathrm{groups} \mathrm{containing} 11,9 \mathrm{and} \mathrm{then} \mathrm{arranging} \\ \mathrm{passengers} \mathrm{in} \mathrm{the} \mathrm{upperdeck} \mathrm{and} \mathrm{lower} \mathrm{deck} \\ {=}^{20}{\mathrm{C}}_9\left(13!\right)(15!) \end{gathered}$$