There are 10 men, 10 women, and 10 rooms. Each person randomly goes into a room. What is the expected number of rooms with at least one man and woman?
Our prof. gave us the following solution however, I'm confused about the probability portion of the answer (especially the $(\frac{9}{10})^{10}$ part):
$$10 \times \left(\!1 - \left(\! \frac{9}{10} \!\right)^{\!\!10}\hspace{1mu}\right)^2$$
 A: The chance that a particular room does not have a specific man is $\frac 9{10}$.  The chance that it does not have any men is $\left (\frac 9{10}\right )^{10}$  The chance that it has at least one man is $1-\left (\frac 9{10}\right )^{10}$  The chance that it has at least one man and at least one woman is $\left(1-\left (\frac 9{10}\right )^{10}\right )^2$  For the expected number of rooms with at least one man and at least one woman, multiply by $10$, getting $$10\left(1-\left (\frac 9{10}\right )^{10}\right )^2$$
A: The result is true. 
It's like the following situation:
You have ten black balls, ten white balls. And there are ten boxes. It's like we randomly throw that balls into the boxes. So for each box $i$, the number of black and white balls that will fall in the box is an event $X_i$. where 
$P(X_i\geq(1,1))=(1-0.9^{10})^2$.
Then we define another variable $Y_i = \begin{cases} 1 & X_i\geq(1,1)\\ 0 &\text{otherwise} \end{cases}$, which means that the result of $box_i$ meet the requirements or not. Then we count the number of success. Note this is not binomial experiments, since $Y_i, Y_j, i\neq j$ are not independent. But this does not matter. We just want to know the expectation of $\mathbb{E}(\sum_i Y_i)$, which is exactly the answer. 
