8. Given the set $A=\{1,2, \cdots, 104\}, S$ is a subset of $A$. If $x \in S$, and at the same time $x-1 \notin S$ and $x+1 \notin S$, then $x$ is called an "isolated point" of $S$. The number of all 5-element subsets of $A$ with no "isolated points" is . $\qquad$
8. 10000 Detailed Explanation: Since the 5-element subset of $A$ has no isolated points, $S$ can only have the following 3 cases. (1) $S=$ $\{a, a+1, a+2, a+3, a+4\}$, where $1 \leqslant a \leqslant 100$. The number of $S$ is 100. (2) $S=\{a, a+1, a+2, b, b+1\}$, where $1 \leqslant a<b-3 \leqslant 100$. The number of $S$ is $C_{100}^{2}$. (3) $S=\{a, a+1, b, b+1, b+2\}$, where $1 \leqslant a<b-2 \leqslant$ 100. The number of $S$ is $C_{100}^{2}$. Therefore, the number of 5-element subsets of $A$ with no isolated points is $100+2 C_{100}^{2}=10000$.