1. **Problem statement:** We need to find the number of different amino-acid sequences possible for an octapeptide containing four of one amino acid, two of another, and two of a third. 2. **Formula used:** This is a permutation problem with repeated elements. The number of distinct sequences of $n$ items where there are groups of identical items is given by the multinomial formula: $$\text{Number of sequences} = \frac{n!}{n_1! \times n_2! \times n_3!}$$ where $n$ is the total number of items, and $n_1, n_2, n_3$ are the counts of each repeated group. 3. **Apply the formula:** Here, $n=8$ (octapeptide), with $n_1=4$, $n_2=2$, and $n_3=2$. $$\text{Number of sequences} = \frac{8!}{4! \times 2! \times 2!}$$ 4. **Calculate factorials:** - $8! = 40320$ - $4! = 24$ - $2! = 2$ 5. **Substitute and simplify:** $$\frac{40320}{24 \times 2 \times 2} = \frac{40320}{96} = 420$$ 6. **Interpretation:** There are 420 different possible amino-acid sequences for the given composition. **Final answer:** $$\boxed{420}$$