In how many different ways can I sort $X$ unique items into $Y$ boxes so that each box has the same number of items? The boxes are all the same.
Specifically, I'm looking for where $X=28$ and $Y=4$.
I tried googling it but I can't find the right phrase.
 A: The answer is $${28!\over7!7!7!7!}.$$  Put the items in a line in $28!$ ways, and put the first $7$ items in the first box, the next $7$ in the second box, and so on.  Since the order the items are placed into a box doesn't matter, we have to divide by $7!$ once for each box. 
A: @saulspatz has already given the answer for your specific case. This answer works perfectly, since your $X$ is divisible by $Y$. But what if that wasn't the case? We'd have to leave some items outside of the boxes for sure. Let me try this task for the general case:
Let $X, Y \in \mathbb{N}$, $X$ being the amount of items, and $Y$ being the amount of boxes. The maximum amount of items we can insert per box is $\left\lfloor \frac{X}{Y} \right\rfloor$. I'll assume that you are free to choose whether you want to actually insert the maximum number of items into each box, or whether you want to settle for a smaller number. (i.e. if max number is 3, you could also just insert 2, 1 or 0 each).
Now, let's first assume we've settled for a number of items we want to insert per box. Call this number $i$. We can simply use the idea of @saulspatz's solution; line the items up in a row, put the first $i$ items into one box, the next $i$ items into the next, and so on. 
Once all boxes are filled, you simply discard the leftover items (imagine it as a "leftovers" box). As we don't care about the order of the items in the "leftovers" box, we'll have to divide by the faculty of the number of leftovers (which is $X\ \mathrm{mod}\ i$) as well. Unfortunately, this can lead to division by $0$ if there are no leftovers, so we'll need to introduce a function that maps to the modulo if it is not zero, and to 1 otherwise: 
$$f_d : \mathbb{N}_0 \rightarrow \mathbb{N},\ f_d: x \mapsto \begin{cases}
x\ \mathrm{mod}\ d \neq 0 & x\ \mathrm{mod}\ d \\
otherwise & 1
\end{cases}$$
So we get:
$$
n = \frac{X!}{(i!)^Y \cdot (f_i(X))!}
$$
If we assume unlabeled boxes, we'll also have to divide by $Y!$ as you've already discussed:
$$
n = \frac{X!}{(i!)^Y \cdot Y! \cdot (f_i(X))!}
$$
Now all we have to do is iterate this through all possibilities for $i$ and sum it up as stated by the sum rule. So, our final number of possibilites $n$ would be:
$$
n = \sum_{i=0}^{\left\lfloor \frac{X}{Y}\right\rfloor} \frac{X!}{(i!)^Y \cdot Y! \cdot (f_i(X))!}
$$
