Nine people in an elevator Assume there are nine people who enter a nine-floor building. Each person can exit on each floor with the same probability. What is the probability that the elevator stops on each floor if:

*

*the building has basement

*the building does not have basement

My aproach.
(A) Building has basement
In this case the setting seems to be simple. Let's denote with $f_i$ the floor chosen by $i^\text{th}$ person. Then we could consider all possible values of tuple $(f_1, f_2, ..., f_9)$ with $f_i \in \{1, ..., 9\}, i=\overline{1,9}$. There are $9^9$ possible values in total and $9!$ possible values where all $f_i$'s are different. Thus, the final probability is $$p=\frac{9!}{9^9}$$
(B) Building without basement
This case is much more difficult for me. Let's consider the same tuple $(f_1, f_2, ..., f_9)$ but with $f_i \in \{2, ..., 9\}, i=\overline{1,9}$. The number of possible values is $8^9$ now. Next, we need to count the number of possible values of the tuple such that it contains all numbers from $\{2, ..., 9\}$.
Eight of of nine people could pick 8 different floor in $8!$ ways while the last one could pick any of 8 floors. Thus the total number of interest is $8\cdot8!$. And final probability is $$p=\frac{8\cdot8!}{8^9}=\frac{8!}{8^8}$$ which contradicts intuitition since it is equal to the probability for the problem (A) but for 8 people. But adding extra person should have increased the probability.
Help much appreciated in validating part (A) and helping with solving part (B).
 A: Case A and B are not clearly stated in the question. Assuming in a building with basement, all $9$ people enter at the basement and there are $9$ floors where they can exit, then your answer for case (A) is correct.
Case (B): If there is no basement, you seem to be  assuming everyone enters on the ground/first floor and there are $8$ floors where they can exit. In that case,
Exactly two people will exit on the same floor and one person each on remaining $7$ floors. So we choose the floor where we have two people exit and we choose two people who exit together and then remaining $7$ people exit on remaining $7$ floors in $7!$ ways. So the probability is,
$\displaystyle P = 8 \cdot {9 \choose 2} \cdot 7! \ / \ 8^9 = \frac{9!}{2! \cdot 8^8}$
A: Part (B) is analogous to an eight faced die thrown $9$ times, and the valid ways can be expressed as the product of two multinomial coefficients, viz $\binom{9}{2,1,1,1,1,1,1,1}\binom{8}{1,7} = \frac{9!}{2!}\frac{8!}{7!}$
You can now compute the probability
