Question

Three persons enter a lift at the ground floor. The lift will go up to the 10th floor. The number of ways in which the three persons can exit the lift at three different floors, if the lift does not stop at the 1st, 2nd and 3rd floors, is equal to _______.

Hint:

In problems involving people exiting at different floors, always multiply combinations of floors by permutations of people.

Answer 1

Correct Answer: 210

Step 1: Identify allowed floors The lift goes from ground floor to 10th floor but does not stop at: \[ 1^\text{st},\ 2^\text{nd},\ 3^\text{rd}\text{ floors} \] Hence, possible exit floors are: \[ 4,5,6,7,8,9,10 \] Total available floors \(=7\).
Step 2: Choose distinct floors Three persons must exit at three different floors. Number of ways to choose 3 distinct floors from 7: \[ {}^7C_3=35 \]
Step 3: Assign persons to floors The three persons are distinct, so they can be arranged among the chosen floors in: \[ 3!=6\ \text{ways} \]
Step 4: Total number of ways \[ \text{Total ways}= {}^7C_3 \times 3! =35\times6 =210 \] Final Answer: \[ \boxed{210} \]