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
\[ \boxed{2520} \]
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