Elevator Wait Time Question: A building contains one elevator which can access each floor numbered $0$ through $m>0$. It picks up $n\le m$ passengers in the lobby (floor 0). If it takes the elevator $\alpha$ seconds at each floor it stops at to open/shut the doors, and $\beta$ seconds to travel between each consecutive floor, what is the expected time to drop off all passengers and return to the lobby.  Also nice would be to find the variance of this time. We can assume that no new passengers get on and each floor is equally likely for each passenger to choose.
My approach: We fix $n\ge 1$ and consider the floor choices for each passenger $\{F_1,F_2,\dots, F_n\}\subset \{1,2,\dots,m\}$ with $P(F_i=k)=\frac{1}{m}$ for all $1\le k \le m$
Now let $N=\max (F_1,F_2,\dots , F_n)$ be the top floor reached.
$P(N\le k) = P(F_i \le k,  \forall i)=[P(F_1\le k)]^n=\left( \frac{k}{m}\right)^n$
So $P(N=k)= P(N\le k)-P(N\le k-1)=\left(\frac{k}{m}\right)^n - \left( \frac{k-1}{m}\right)^n=\frac{k^n}{m^n}\left(1-\left(1-\frac{1}{k}\right)^n\right)$
So, given $N=k$, we want to find $X$, the number of stops made:
WLOG we can assume $F_1\le F_2\le \cdots \le F_{n-1}\le F_n=k$
so $P(F_i=\ell | N=k)=\frac{1}{k}$
Here is where I am not so sure:
$P(X=j|N=k)$ doesnt seem to yield anything nice. Perhaps I have travelled down the wrong path.
This question was in a document recommended for me to inspire bright highschool students, so I feel like there should be a simple answer, or at least a way to justify some calculations being omitted. any light you can shed would be appreciated.

Outline: The total wait time $W$ is the sum of the total open/close time $X$ and the travel time $Y$. So by the Linearity of Expectation the expectation of $W$ is the sum of $E(X)$ and $E(Y)$.
I don't know whether we are to count the open/close time at floor $0$. Let's decide not to. For floors $1$ to $m$, let $X_i=1$ if we stop at Floor $i$, and $X_i=0$ otherwise. Then $X=\alpha(X_1+\cdots+X_m)$, and by the linearity of expectation
$$E(X)=\alpha\sum_1^m E(X_i).$$
The probability we stop at floor $i$ is $1$ minus the probability we do not stop. The probability we do not stop is $\left(\frac{m-1}{m}\right)^n$. Putting things together, we find that
$$E(X)=\alpha m\left(1-\left(\frac{m-1}{m}\right)^n\right).$$
The expectation of $Y$ is $2\beta$ times the expected maximum height reached. It looks as if you can handle that part.
