Find all eight-digit numbers formed only with the digits 1, 2, 3, and 4, each of them twice, such that:

(a) there is one digit between the two digits 1;

(b) there are two digits between the two digits 2;

(c) there are three digits between the two digits 3 and

(d) there are four digits between the two digits 4.
The given conditions imply that the numbers must satisfy all the following conditions.

$$
\cdots \underline{1}-1 \cdots
$$

$$
\cdots \underline{2}--\underline{2} \cdots
$$

$$
\cdots \underline{3}--\underline{3} \cdots
$$

$$
\cdots \underline{4}---\underline{4} \cdots
$$

| case A: | $\underline{4}--\underline{3}-\underline{4}-\underline{3}$ | or | $\underline{4}-\underline{3}-\underline{4} \underline{3}-$ |
| :---: | :---: | :---: | :---: |
| case B: | $3 \underline{4}-$ | or | -43 |
| case C: | $\underline{3}-\underline{4}-\underline{3}--\frac{4}{4}$ | or | $-\underline{3} \underline{4}-\underline{3}-\underline{4}$ |

Let's study the possible positions of the two digits 4 in an eight-digit number. According to (iv), there are only three possibilities:

$$
\begin{array}{ll}
\text { case A: } & \underline{4}-\underline{4}--\underline{4}-\frac{4}{4} \\
\text { case B: } & -\underline{4}---\underline{4} \\
\text { case C: } & --4---4
\end{array}
$$

In each of these cases, there are two possibilities for placing the digits 3:

In the attempt to place the digits 1 and 2, we realize that the two possibilities of case B are impossible, as well as the first possibilities of cases A and C. The only cases that lead to solutions of the problem are the second possibilities of cases A and C, which lead to the two unique solutions

$$
41312432 \text { and } 23421314
$$