Let $i_{1}, i_{2}, \ldots i_{n}$ be any permutation of the elements $1,2, \ldots n$ and consider the system of inequalities

$$
\begin{aligned}
& x_{1}>x_{2}>\ldots>x_{n-1}>x_{n} \ldots \\
& \left|x_{i_{1}}\right|>\left|x_{i_{2}}\right|>\ldots>\left|x_{i_{n-1}}\right|>\left|x_{i_{n}}\right| \ldots
\end{aligned}
$$

Determine how many permutations $i_{1}, i_{2}, \ldots, i_{n}$ exist for which this system is solvable.
We will show that

$$
i_{1}=k, \text { if } k=2,3, \ldots(n-2),(n-1)
$$

is not possible, that is, for permutations of the elements $1,2, \ldots n$ where the first position is occupied by $2,3, \ldots(n-2),(n-1)$, the system consisting of $1a$) and $1b$) cannot be solved.

If indeed $i_{1}=k[$ where $k=2,3, \ldots(n-2),(n-1)]$, and the system in question had a solution, then according to $\left.1 b\right)$, $\left|x_{k}\right|>\left|x_{k-j}\right|$ and according to $\left.1 a\right)$, $x_{k-j}>x_{k}$ is only possible if $x_{k}x_{k}>x_{k+l}^{2}$ while $\left.1 b\right)$ states $\left|x_{k}\right|>\left|x_{k+l}\right|$, 

thus we arrive at a contradiction.

Therefore, $i_{1}$ can only be 1 or $n$, in other words, the smallest or the largest of the elements $1, 2, \ldots n$ (in natural order). It is already clear that if we place one of the above in the position of $i_{1}$, then the

$$
i_{2}, i_{3}, \ldots i_{n}
$$

can only be a permutation where the smallest or the largest of the remaining elements, excluding the one placed in the position of $i_{1}$, can appear in the position of $i_{2}$. If, therefore, the number of suitable permutations standing in the positions

$$
i_{2}, i_{3}, \ldots i_{n}
$$

is $p_{n-1}$, and the number of suitable permutations formed in the positions

$$
i_{1}, i_{2}, i_{3}, \ldots i_{n}
$$

is $p_{n}$, then

$$
p_{n}=2 p_{n-1}
$$

Since $p_{2}=2$, that is, the system

$$
x_{1}>x_{2} \text { and }\left|x_{i_{1}}\right|>\left|x_{i_{2}}\right|
$$

is solvable for the 2 permutations of the elements 1,2,

$$
p_{3}=2 p_{2}=2^{2}, \quad p_{4}=2^{3}, \ldots \quad p_{n}=2^{n-1}
$$

Thus, the number of permutations of the elements $1,2, \ldots n$ for which the system $1a$) - $1b$) is solvable is $2^{n-1}$. For example, let $n=5$. The system

$$
\begin{gathered}
x_{1}>x_{2}>x_{3}>x_{4}>x_{5} \ldots \\
\left|x_{i_{1}}\right|>\left|x_{i_{2}}\right|>\left|x_{i_{3}}\right|>\left|x_{i_{4}}\right|>\left|x_{i_{5}}\right| \ldots
\end{gathered}
$$

has solutions for the following permutations of the elements $1,2,3,4,5$:

| 12345 | 15234 | 51234 | 54123 |
| :--- | :--- | :--- | :--- |
| 12354 | 15243 | 51243 | 54132 |
| 12534 | 15423 | 51423 | 54312 |
| 12543 | 15432 | 51432 | 54321 |

${ }^{1} j=1,2, \ldots(k-1)$.

$2 l=1,2, \ldots(n-k)$