Question

In a $12$-storey house, ten people enter a lift cabin. It is known that they will leave the lift in pre-decided groups of $2,3,$, and $5$ people at different storeys. The number of ways they can do so if the lift does not stop at the second storey is
1. $78$
2. $112$
3. $720$
4. $132$

Answer 1

Hint: In this question, we have to find the number of ways that even can happen. We’ll use a combination. According to the question the lift is not stopping at the second floor which means the floor would not be considered in the selection and they are entering from the ground floor. Therefore, the number of possible floors on which the lift will stop is ten. Also, people are leaving the lift in the group of $2,3,$ and $5$so we’ll not make any change.

Formula Used:
Combination formula –
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)......$

Complete step by step Solution:
Given that,
The house is of $12$ storey where the lift will not stop at the second storey and they will enter in a lift at the ground floor, it implies that only $10$ floors will be considered.
Now, only three floors will be used
Therefore, total combinations$ = {}^{10}{C_3}$
There are three groups who have decided to leave the lift, the combination is $3!$
Hence, the total number of ways the lift does not stop at the second storey is
$ = {}^{10}{C_3} \times 3!$
$ = \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} \times 3!$
$ = \dfrac{{10!}}{{7!}}$
$ = \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!}}$
$ = 720$

Hence, the correct option is (3).

Note:In such questions, always read the workings carefully. Like here it is written that we have to find the ways that the lift will not stop at the second floor so we’ll consider that floor only and will count eleven floors while calculating. But here we have to count ten floors only because from where the lift will start on that floor also lift will not stop.