Question

A double-decker can accommodate $20$ passenger, $7$ in the lower deck and $13$ in the upper deck. Find the number of ways the passengers can be accommodated if $5$ want to sit only in the lower deck and $8$ want to sit only in the upper deck.

Answer 1

Remember you don’t have to arrange the passengers who are already sitting or have a preference. Arrange the remaining passengers using combinations in the remaining seats in the lower or upper deck.

Complete step-by-step answer:
Let’s first try to analyse the question properly. In the given problem, there are total $20$ passengers that are to be arranged in $20$ seats of a double-decker that has $7$ seats in the lower deck and $13$ in the upper deck. But we also have a condition that, $5$ passenger should be arranged only in the lower deck and $8$ passengers should be arranged only in the upper deck.
So basically, $5$ passengers in $7$ seats of the lower deck are already arranged then $8$ passengers in $13$ seats of the upper deck are already arranged. And then remaining $7$ passengers in $7$ remaining seats.
Now we can start arranging passengers using combinations step by step. But for that, you need to know the idea of combination. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter.
$\Rightarrow$ If $'n'$ is the number of things to choose from, and we choose $'r'$ of them, no repetition, the order doesn't matter; then we can represent then as: {}^n{C_r} = \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Out of $7$ remaining passengers, we have to arrange them in remaining $2$ seats in the lower deck and in $5$ seats in the upper deck.
$\Rightarrow$ We can arrange remaining passengers in the upper deck then after those remaining two passengers will be arranged in remaining two seats by itself $\Rightarrow {}^7{C_5} = \dfrac{{7!}}{{5!\left( {7 - 5} \right)!}}$
Therefore, the number of required ways$= \dfrac{{7!}}{{5!\left( {7 - 5} \right)!}} = \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}}{{1 \times 2 \times 3 \times 4 \times 5 \times 1 \times 2}} = \dfrac{{6 \times 7}}{{1 \times 2}} = 21$
Hence, the number of ways to arrange the passengers as required is $21$.

Note: Try to visualize the given condition properly before starting a solution. An alternative approach can be taken by arranging the remaining passengers in the lower deck first, i.e. by using${}^7{C_2}$. But you will find the same answer since ${}^n{C_r} = {}^n{C_{n - r}}$