Covariance of 2 sequences of random variables We have a random number generator that generates random integers independently uniformly distributed from 1 to 5, inclusive. This random number generator is used to generate a sequence on $n$ independent random integers.
Let $X$ = number of $1$’s generated in the sequence of $n$ integers. Let $Y$ = number of $5$’s generated in the sequence of $n$ integers.
(Hint: it might be useful to define indicator variables denoting whether the $i$-th integer generated is a 1, and likewise for whether the $i$-th integer generated is a 5.)
What is $\text{cov}(X, Y)$?
 A: As a response to a self-study question, this answer aims to share a problem solving process rather than merely stating an answer. 

There doesn't seem to be anything special about ones and fives: because "one" and "five" are just different labels for two of the five possible outcomes, the joint distribution of $(X,Y)$ will be identical to the joint distribution of a corresponding $X$ and $Y$ if we were to replace $1$ and $5$ by any two distinct values.
With this in mind, let's apply the hint by defining $Z^{(j)}_i = 1$ when the $i^\text{th}$ random integer has the value $j$ (and otherwise $Z^{(j)}_i=0$).  Thus the number of $j$s appearing in the sequence 
is
$$Z^{(j)} = \sum_{i=1}^n Z^{(j)}_i.$$
Indeed, let's generalize to the case $5 = M$, where $M$ is some positive integer.  In this notation $X = Z^{(1)}$ and $Y=Z^{(M)}$.
The question asks about covariances.  These depend on the first two moments. 
Having already observed that $1$ and $5$ are not so special, we might be inspired to consider a beautifully symmetric quantity
$$Z = Z^{(1)} + Z^{(2)} + \ldots + Z^{(M)}.$$
As a random variable it's rather uninteresting: by totaling all $M$ counts, it must invariably equal $n$ But as an example of its usefulness consider its first moment,
$$n = \mathbb{E}(n) = \mathbb{E}(Z) = \mathbb{E}\left(Z^{(1)} + Z^{(2)} + \ldots + Z^{(M)}\right) \\= \mathbb{E}(Z^{(1)}) + \mathbb{E}(Z^{(2)}) + \cdots + \mathbb{E}(Z^{(M)})$$
The reasoning (about distributional equivalence) at the outset implies all $M$ of these counts must have equal expectations.  Therefore
$$\frac{n}{M} = \mathbb{E}(Z^{(1)}) = \mathbb{E}(Z^{(2)}) = \cdots = \mathbb{E}(Z^{(M)}).$$
That was easy!
Let's try a second-order example by computing the variance of $Z$. This obviously is zero, because $Z$ does not vary at all, whence
$$0 = \text{Var}(Z) = \text{Var}\left(Z^{(1)} + Z^{(2)} + \ldots + Z^{(M)}\right).$$
Algebraic rules for computing variances allow this to be expressed as sums of the variances of the $Z^{(j)}$ and the covariances of the $Z^{(j)}$ and $Z^{(l)}$ (for $j\ne l$).  That is promising: if we could find the variances, we could solve for the common covariance, which is the goal of this exercise.
As a final hint, note that each $Z^{(j)}$ has a Binomial$(n,1/M)$ distribution.
