A double-decker bus has $ 5 $ empty seats in the upstairs and $ 5 $ empty seats in the downstairs. $ 10 $ people aboard the bus of which $ 2 $ are old people and $ 3 $ are children. The children refuse to take seats down stairs while old people insist to stay down stairs. In how many different arrangements can the $ 10 $ people take seats in the bus.
A. $ 144000 $
B. $ 72000 $
C. $ 36000 $
D. $ 18000 $

Hint: Since the children and old people have mentioned their preferences of seats, we will give priority to old men and children. Firstly, we will select $ 3 $ seats among $ 5 $ seats upstairs for children and then we will select $ 2 $ seats among $ 5 $ seats for old people downstairs. Finally, we will find out the total number of ways for the seating arrangement of $ 10 $ people in the bus.

Complete step-by-step answer:
Given: There are total $ 10 $ people,( $ 2 $ old people, $ 3 $ children), who aboard a double decker bus
Step $ 1 $ : There are $ 5 $ seats upstairs and $ 5 $ seat downstairs.
Selection of $ 3 $ seats in upstairs for children and the arrange children there
$ { = ^5}{C_3} \times 3! $
Step $ 2 $ : There are $ 5 $ seats downstairs.
Selection of $ 2 $ seats in downstairs for old people and arrange them there
$ { = ^5}{C_2} \times 2! $
Step $ 3 $ : Total number of ways for the seating arrangement of $ 10 $ people are
 $
  { = ^5}{C_3} \times 3!\;{ \times ^5}{C_2} \times 2! \times 5! \\
   = 144000 \\
  $
So, the correct answer is “Option A”.

Note: In these types of questions we always give the priority to those who are either always included or excluded. Also, not to be confused between permutation and combination. The various ways in which objects from a set may be selected, generally without replacement, to form subsets, This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.