# Find the probability that the $8$ people will descend on different floors.

An elevator in a building with $$10$$ floors and a ground floor is approached on the ground floor for $$8$$ people:

• Each person chooses at random ( with uniform probability ) the floor on which they will get off the elevator.
• Find the probability that the $$8$$ people will descend on different floors.

Idea: I think the solution is $$\displaystyle\frac{10\times9\times8\times7\times6\times5\times4\times3}{10^8}$$. Is this correct ?.

• Yes, your answer is correct. Commented Aug 16, 2020 at 16:06

The solution will be: $${\text{No. of permutations where each person descends on a different floor}\over\text{Total no. of permutations}}$$
• Number of permutations where each person descends on a different floor = $$\frac{10!}{2!}$$
$$\text{Person}_1$$ chooses from 10 floors, $$\text{person}_2$$ chooses from remaining 9 floors, $$\text{person}_3$$ chooses from remaining 8 floors, $$\ldots$$. This equals $$10*9*8* ... *3$$.
• Total no. of permutations = $$10^{8}$$
$$\text{Person}_1$$ chooses from 10 floors, $$\text{person}_2$$ also chooses from 10 floors, $$\text{person}_3$$ also chooses from 10, $$\ldots$$. This equals $$10^{8}$$.
So the solution = $${10!\over\text10^{8}\times2!}$$