A standard six-sided die is rolled $8$ times. You are told that among the rolls, there was two $1$'s, two $2$'s, two $3$'s, and two $4$s. How many possible sequences of rolls could there have been? (For example, $2,$ $1,$ $3,$ $4,$ $2,$ $1$, $3$, $4$ is one possible sequence.)

To find the number of possible sequences of rolls when rolling a six-sided die \(8\) times, we must account for the specified occurrences: we have two \(1\)s, two \(2\)s, two \(3\)s, and two \(4\)s.

Since the rolls total \(8\), and we have fixed categories for the outcomes, we can represent our sequence using a multiset. We can express this situation with the following multiset representation:

\[
\{1, 1, 2, 2, 3, 3, 4, 4\}
\]

Next, we can compute the total number of distinct sequences of this multiset. The formula for the number of distinct permutations of a multiset is given by the expression:

\[
\frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots}
\]

Where \(n\) is the total number of elements, and \(n_1, n_2, n_3, \ldots\) are the frequencies of each unique element.

In our case:
- The total number of elements \(n = 8\) (the total number of rolls).
- The frequencies of the numbers are: \(n_1 = 2\) (for \(1\)), \(n_2 = 2\) (for \(2\)), \(n_3 = 2\) (for \(3\)), and \(n_4 = 2\) (for \(4\)).

Thus, we have:

\[
\text{Number of sequences} = \frac{8!}{2! \times 2! \times 2! \times 2!}
\]

Calculating \(8!\):

\[
8! = 40320
\]

Now calculating \(2!\):

\[
2! = 2
\]

Then calculating \(2! \times 2! \times 2! \times 2!\):

\[
2! \times 2! \times 2! \times 2! = 2 \times 2 \times 2 \times 2 = 16
\]

Now we compute the number of distinct arrangements:

\[
\frac{8!}{2! \times 2! \times 2! \times 2!} = \frac{40320}{16} = 2520
\]

Thus, the total number of possible sequences of rolls is

\[
\boxed{2520}
\]