In a building, an elevator starts from the ground floor (that is, 0th floor) with $10$ passengers and stops at every floor until the 15th floor which is the topmost floor. No new passengers enter the elevator on any floor from the 1st through the 15th floor. If all the passengers got off the elevator then the probability that at most one passenger got off at each floor is

Total number of passengers is $10$.

Total number of floors where passengers can get off is $15$ (from the 1st to the 15th floor).

Since each passenger can get off at any of the $15$ floors, the total number of ways for the $10$ passengers to get off is $15^{10}$.

The condition that at most one passenger gets off at each floor implies that all $10$ passengers must get off at $10$ distinct floors.

The number of ways to choose $10$ floors out of $15$ is $^{15}C_{10}$.

The number of ways the $10$ passengers can be arranged in these $10$ chosen floors is $10!$.

The number of favorable ways is $^{15}C_{10} \times 10! = \dfrac{15!}{10!5!} \times 10! = \dfrac{15!}{5!}$.

The required probability is the ratio of the number of favorable ways to the total number of ways, which is $\dfrac{15!}{5! \times 15^{10}}$.