Question

There are $10$ seats in a double-decker bus, $6$ in the lower deck, and $4$ on the upper deck. Ten passengers board the bus, of them $3$ refuse to go to the upper deck and $2$ insist on going up. The number of ways in which the passengers can be accommodated is______. (Assume all seats to be duly numbered)

Answer 1

Hint: There are ten seats in the bus and ten passengers board the bus in which $3$ refuse to go up and $2$ insist on going up. So they have to be given the seats according to their choice but the rest of the $5$ passengers can be seated in any order. So we will use the formula of combination to solve this question which is ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r! n - r!}}}}$

Complete step by step answer:

Given, the total number of seats is $10$.The number of seats in the lower deck is $6$and in the upper deck is $4$.The number of passengers boarding the bus is also $10$.Now out of 10 passengers, $3$ refuses to go up and $2$ insists on going up. We know that the number of ways to select r things from n things $ = {}^{\text{n}}{{\text{C}}_{\text{r}}}$ and the number of ways the n seats can be filled is (n!). So the number of ways in which 2 out of 4 upper deck seats are selected and filled by the $2$ people who insist on going up $ = {}^4{{\text{C}}_2} \times 2!$
Similarly the number of ways in which the 3 out of 6 seats can be selected for the people, who refuse to go up, and filled by the 3 people $ = {}^6{{\text{C}}_3} \times 3!$. In this way 5 seats are filled, now only 5 seats and 5 people are left. So the number of ways in which $5$ seats can be filled by $5$people$ = 5!$
So the number of ways in which the passengers can be accommodated $ = {}^4{{\text{C}}_2} \times 2! \times {}^6{{\text{C}}_3} \times 3! \times 5!$
The formula of combination ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r! n - r!}}}}$ where n is the total number of elements in a set and r are the number of elements to be selected, (!) is the sign of factorial. Here,${\text{n}}! = {\text{n}} \times \left( {{\text{n - 1}}} \right) \times ... \times 3 \times 2 \times 1 = {\text{n}} \times \left( {{\text{n}} - 1} \right)!$
On putting the given value in the formula we get-
The number of ways in which the passengers can be accommodated $ = \dfrac{{4!}}{{2!4 - 2!}} \times 2! \times \dfrac{{6!}}{{3!6 - 3!}} \times 3! \times 5! = \dfrac{{4!}}{{2!}} \times \dfrac{{6!}}{{3!}} \times 5!$
On simplifying we get,
$ \Rightarrow \dfrac{{4 \times 3 \times 2!}}{{2!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5! = 12 \times 120 \times 120 = 172,800$
Answer-The number of ways the passengers can be accommodated is $172,800$.

Note: Combinations can be confused with permutations. However, in permutations, the order of the selected items is essential. A combination determines the number of possible arrangements in a collection of items where the order of the selection does not matter. The items can be selected in any order.