Let [tex]X_i[/tex] be the random variable indicating whether the elevator does not stop at floor [tex]i[/tex], with
[tex]X_i=\begin{cases}1&\text{if the elevator does not stop at floor }i\\0&\text{otherwise}\end{cases}[/tex]
Let [tex]Y[/tex] be the random variable representing the number of floors at which the elevator does not stop. Then
[tex]Y=X_1+X_2+\cdots+X_{n-1}+X_n[/tex]
We want to find [tex]\mathrm{Var}(Y)[/tex]. By definition,
[tex]\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2[/tex]
As stated in the question, there is a [tex]\dfrac1n[/tex] probability that any one person will get off at floor [tex]n[/tex] (here, [tex]n[/tex] refers to any of the [tex]n[/tex] total floors, not just the top floor). Then the probability that a person will not get off at floor [tex]n[/tex] is [tex]1-\dfrac1n[/tex]. There are [tex]m[/tex] people in the elevator, so the probability that not a single one gets off at floor [tex]n[/tex] is [tex]\left(1-\dfrac1n\right)^m[/tex].
So,
[tex]\mathbb P(X_i=x)\begin{cases}\left(1-\dfrac1n\right)^m&\text{for }x=1\\\\1-\left(1-\dfrac1n\right)^m&\text{for }x=0\end{cases}[/tex]
which means
[tex]\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)[/tex]
[tex]\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m[/tex]
and
[tex]\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i<j}X_iX_j\right]=\displaystyle\sum_{i=1}^n\mathbb E[{X_i}^2]+2\sum_{1\le i<j}\mathbb E[X_iX_j][/tex]
Computing [tex]\mathbb E[{X_i}^2][/tex] is trivial since it's the same as [tex]\mathbb E[X_i][/tex]. (Do you see why?)
Next, we want to find the expected value of the following random variable, when [tex]i\neq j[/tex]:
[tex]X_iX_j=\begin{cases}1&\text{if }X_i=1\text{ and }X_j=1\\0&\text{otherwise}\end{cases}[/tex]
If [tex]X_iX_j=0[/tex], we don't care; when we compute [tex]\mathbb E[X_iX_j][/tex], the contributing terms will vanish. We only want to see what happens when both floors are not visited.
[tex]\mathbb P(X_iX_j=1)=\left(1-\dfrac2n\right)^m[/tex]
[tex]\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m[/tex]
[tex]\implies2\displaystyle\sum_{1\le i<j}\mathbb E[X_iX_j]=2n(n-1)\left(1-\dfrac2n\right)^m[/tex]
where we multiply by [tex]n(n-1)[/tex] because that's how many ways there are of choosing indices [tex]i,j[/tex] for [tex]X_iX_j[/tex] such that [tex]1\le i<j[/tex].
So,
[tex]\mathrm{Var}[Y]=n\left(1-\dfrac1n\right)^m+2n(n-1)\left(1-\dfrac2n\right)^m-n^2\left(1-\dfrac1n\right)^{2m}[/tex]
