Question

In a 12 storey house, 10 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 up to the second storey is

• A. 120
• B. 78
• C. 132
• D. 720

Hint:

In combinatorics problems where multiple groups need to be arranged, use the formula for permutations of multi-sets: \[ \frac{n!}{k_1! k_2! \cdots k_r!} \] where \( n \) is the total number of items, and \( k_1, k_2, \dots \) are the sizes of the individual groups.

Answer 1

The Correct Option is D

we are given a group of 10 people that need to exit in groups of 2, 3, and 5 at different storeys. The problem can be broken down as follows:
1. First, we need to assign 3 groups of people (2, 3, and 5) to different storeys. This can be done in \( \frac{10!}{2!3!5!} \) ways, as we are dividing the 10 people into distinct groups of 2, 3, and 5.
2. Second, for each of the groups (2 people, 3 people, and 5 people), we can choose the storeys at which they will exit. The number of ways to arrange 3 groups in 3 different storeys is simply \(3! = 6\).
Thus, the total number of ways the 10 people can exit the lift is calculated by multiplying these two values: \[ \frac{10!}{2!3!5!} \times 3! = \frac{10!}{2!3!5!} \times 6 = 720 \]