8. Suppose $A=\{1,2, \ldots, 20\}$. Call $B$ a visionary set of $A$ if $B \subseteq A, B$ contains at least one even integer, and $|B| \in B$, where $|B|$ is the cardinality of set $B$. How many visionary sets does $A$ have?
Answer: $2^{19}-2^{8}$ or $2^{19}-256$
Solution: Ignoring the constraint that all visionary sets must contain at least one even integer, we count a total of
$$
\sum_{i=0}^{19}\binom{19}{i}=2^{19}
$$
sets. From this total, we subtract the number of sets that only contain odd numbers. Evidently, $|B|$ has to be odd. Since there are only 10 odd numbers in $A,|B|$ can only be $1,3,5,7$, or 9 . The total number of instances when this happens is
$$
\sum_{i=0}^{4}\binom{9}{2 i}=\frac{1}{2} \sum_{i=0}^{9}\binom{9}{i}=\frac{1}{2}\left(2^{9}\right)=2^{8}
$$
sets. Therefore, there are a total of $2^{19}-2^{8}$ visionary sets.