A lift is going up to the \(10^{\text{th}}\) floor.
Number of ways in which \(3\) people can exit the lift at {three different floors,
if the lift will not stop at I$^{\text{st}}$, II$^{\text{nd}}$ and III$^{\text{rd}}$ floors, is:
}
Show Hint
When {people are distinct} and {floors are distinct},
use {permutations} rather than combinations.
To solve this problem, we need to determine the number of ways 3 people can exit a lift at three different floors, given that the lift does not stop at the 1st, 2nd, and 3rd floors.
The lift operates from the 4th to the 10th floor, which means there are a total of 10 - 3 = 7 floors available for stopping (4th to 10th floors).
Since the three people must exit at different floors, we need to choose 3 different floors from these 7 available floors.
The number of ways to choose 3 different floors from 7 is given by the combination formula: {}^nC_r, where n is the total number of options, and r is the number of selections being made. Here, n = 7 and r = 3.
Therefore, the calculation is:
{}^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
Now, for each selection of 3 different floors, there are 3! (factorial of 3) ways to allocate the floors among the 3 people (as each person exits at a distinct floor). The calculation is:
3! = 3 \times 2 \times 1 = 6
Hence, the total number of ways is determined by multiplying the number of ways to choose the floors by the number of distinct arrangements of people:
35 \times 6 = 210
Therefore, the number of ways in which 3 people can exit the lift at three different floors (excluding the 1st, 2nd, and 3rd floors) is 210. Thus, the correct answer is \(210\).
