1.10 $E_{2} \neq \varnothing, E_{2} \cap E_{3} \neq \varnothing, E_{3} \cap E_{4} \neq \varnothing$.  How many such subset groups are there?
Parse: Without considering the conditions, each $x \in\{1,2, \cdots, n\}$ has exactly $2^{4}$ choices, therefore, there are $16^{n}$ sets of $\left(E_{1}, E_{2}, E_{3}, E_{4}\right)$. Among these, the sets of groups where at least one condition is not met (for example, $E_{1} \cap E_{2}=\varnothing$) total $12^{n} \times 3$ (at this time, each element has exactly $3 \times 2^{2}=12$ choices), such groups should be deducted. Then consider the sets of groups where at least two conditions are not met, there are two types, the numbers (as analyzed above) are $10^{n} \times 2$ and $9^{n}$, respectively, which should be added back. Finally, the groups where two conditions are not met should be deducted, the number is $8^{n}$.
In summary, the number of subsets that meet the conditions is $16^{n}-12^{n} \times 3+\left(10^{n} \times 2+9^{n}\right)-8^{n}$.