Five persons entered the lift cabin on the ground floor of an 8-floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first. The probability of all five persons leaving at different floors, is

$\frac{^7C_5×5!}{7^5}$

Besides ground floor, there are 7 floors. Since a person can leave the cabin at any of the seven floors, therefore the total number of ways in which each of the five persons can leave the cabin at any of the 7 floors = $7^5$.

∴ Total number of ways =$7^5$

Five persons can leave the cabin at five different floors in ${^7C_5×5!}$ ways. Therefore,

∴ Favourable number pf ways = ${^7C_5×5!}$

Hence, required probability $=\frac{^7C_5×5!}{7^5}$