Suppose $r\geq 1$ distinct books are distributed at random among $n\geq 3 $ children. Let $X$ be the number of children who do not get any book, and $Y$ be the number of children who get exactly one book. Then find an expression for $\operatorname{Cov}(X,Y)$.
Given $r$, and $n$, $X$ has support $\{0,1,.....,n-1\}$. Calculating its density function is straightforward, $$ \Pr(X=x) = \frac{{n \choose x}(n-x)^{r}}{n^r} $$
I'm not really sure how to calculate a general expression for density of $Y$ and their joint density.
For each $1\leq j \leq n$ take $N_j\sim \text{Binomial}\left(r,\frac{1}{n}\right)$ as the number of books that child $j$ receives.
Put $X_j=1_{\{N_j=0\}}$ and $Y_j=1_{\{N_j=1\}}$. Notice how $X=\sum_{j=1}^nX_j$ and $Y=\sum_{j-1}^nY_j$.
Let's examine $\text{cov}(X_j,Y_j)$ for $j\in\{1,...,n\}$ fixed.
$$\begin{eqnarray*}\text{cov}(X_j,Y_j)&=& \mathbb{E}(X_jY_j)-\mathbb{E}(X_j)\mathbb{E}(Y_j) \\ &=& \mathbb{P}(X_j=1,Y_j=1) - \mathbb{P}(X_j=1)\mathbb{P}(Y_j=1) \\ &=& \mathbb{P}\left(N_j=0,N_j=1\right)-\mathbb{P}(N_j=0)\mathbb{P}(N_j=1) \\ &=& 0-\left[{r \choose 0} \left(\frac{1}{n}\right)^0 \left(1-\frac{1}{n}\right)^{r}\right]\left[{r \choose 1} \left(\frac{1}{n}\right)^1 \left(1-\frac{1}{n}\right)^{r-1}\right] \\ &=& - \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1}\end{eqnarray*}$$
Now let's study $\text{cov}(X_i,Y_j)$ for $i\neq j$:
$$\begin{eqnarray*}\text{cov}(X_i,Y_j) &=& \mathbb{E}(X_iY_j)-\mathbb{E}(X_i)\mathbb{E}(Y_j) \\ &=& \mathbb{P}(X_i=1,Y_j=1)-\mathbb{P}(X_i=1)\mathbb{P}(Y_j=1) \\ &=& \mathbb{P}\left(\text{Child } i \text{ gets no books & child }j\text{ gets one book}\right)- \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1} \\ &=& {r \choose 1}\frac{1}{n} \cdot \left(\frac{n-2}{n}\right)^{r-1}- \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1}\end{eqnarray*}$$
We get $$\begin{eqnarray*}\text{cov}\left(X,Y\right) &=& \text{cov}\left(\sum_{j-1}^n X_j,\sum_{j=1}^n Y_j\right) \\ &=& \sum_{j=1}^n \text{cov}\left(X_j,Y_j\right)+\sum_{i\neq j}\text{cov}(X_i,Y_j) \\ &=& n\text{cov}(X_1,Y_1)+(n^2-n)\text{cov}(X_1,Y_2)\end{eqnarray*}$$ Plug in the formulas we obtained above into $\text{cov}(X_1,Y_1),\text{cov}(X_1,Y_2)$ to finish the job.