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

Question

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 ?.

Answer 1

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 = ${10!\over\text10^{8}\times2!}$

Your answer is right.