Apart from the ground floor, there are 7 floors
Let's find out the total number of ways in which all the five persons can leave the lift at seven different floors
The 1st person can leave the lift in any of the 7 floors (7 ways)
The 2nd person can leave the lift in any of the remaining 6 floors (6 ways)
The 3nd person can leave the lift in any of the remaining 5 floors (5 ways) ...
The 5th person can leave the lift in any of the remaining 3 floors (3 ways)
Total number of ways
n(E)= Total Number of ways in which all the five persons can leave the lift at seven different floors
Now we will find out the total number of ways in which each of the five persons can leave the lift at any of the seven floors
The 1st person can leave the lift in any of the 7 floors (7 ways)
The 2nd person can leave the lift in any of the 7 floors (7 ways)
The 3nd person can leave the lift in any of the 7 floors (7 ways)
...
The 5th person can leave the lift in any of the 7 floors (7 ways)
Total number of ways = 7 × 7 × 7 × 7 × 7 = 7
5i.e., The total number of ways in which each the five persons can leave the lift at any of the seven floors = n(S) = 7
5.
Then, the probability that all the five persons are leaving the lift at different floors