3. Let's consider the set

$$
U=\bigcup_{i=1}^{\infty} \mathcal{P}^{i}(\varnothing)
$$

where $\mathcal{P}(X)$ denotes the set of all subsets of the set $X$, and $\mathcal{P}^{i}(X)$ denotes the expression $\mathcal{P}(\mathcal{P}(\ldots \mathcal{P}(X) \ldots)$, where $\mathcal{P}$ is applied $i$ times.

For a given natural number $n$, we count how many $n$-element subsets $A$ of the set $U$ exist such that $A \subseteq \mathcal{P}(A)$. Is it possible for the result to consist of the digits $2, 0, 1$, and $8$ in some permutation?
3. Let $A=\left\{A_{1}, A_{2}, \ldots, A_{n}\right\} \in U$ be a set such that $A \subset \mathcal{P}(A)$. Its elements $A_{1}, \ldots, A_{n}$ can be numbered so that $A_{i} \not \subset A_{j}$ whenever $i>j$. For each $j=1, \ldots, n$, it holds that $A_{j} \in \mathcal{P}(A)$, i.e., $A_{j} \subset A$, which means that $A_{j}$ is a set whose elements are some of the sets $A_{i}$ for $i<j$. In other words, $A_{1}=\emptyset$ and, for $j \geqslant 2$,

$$
A_{j}=\left\{A_{i} \mid i \in M_{j}\right\} \quad \text { for some set } \quad M_{j} \subset\{1,2, \ldots, j-1\}
$$

where the sets $M_{2}, \ldots, M_{n}$ are mutually distinct.

Define $a_{j}=\sum_{i \in M_{j}} 2^{i-1}$. The number $a_{j}<2^{j-1}$ uniquely determines the set $M_{j}$. Since $A_{i} \subset A_{j}$ implies $a_{i}<a_{j}$, we can assume that $a_{2}<a_{3}<\cdots<a_{n}$. Thus, we have constructed a bijection between the sets $A$ and the sequences $a_{2}<a_{3}<\cdots<a_{n}$ such that $a_{j}<2^{j-1}$ for each $j$.

Let $f(n)$ be the number of all such sequences $a_{2}<\cdots<a_{n}$, and let $f(n, k)$ be the number of those sequences in which $a_{n}=k$. Then $f(n, k)=0$ for $k \geqslant 2^{n-1}$,

$$
f(n, k)=\sum_{i=1}^{k-1} f(n-1, i) \text { for } k<2^{n-1} \quad \text { and } \quad f(n)=\sum_{k=1}^{2^{n-1}-1} f(n, k)
$$

so we can easily compute $f(2)=1, f(3)=2, f(4)=9, f(5)=88$ and $f(6)=1802$. Therefore, the answer is $\partial a$.