5 Let the set $M=\{1,2, \cdots, 10\}$ have $k$ 5-element subsets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the condition: any two elements in $M$ appear in at most two subsets $A_{i}$ and $A_{j}(i \neq j)$, find the maximum value of $k$.


5. Let the number of times an element $i (i=1,2, \cdots, 10)$ in $M$ appears in $A_{1}, A_{2}, \cdots, A_{k}$ be denoted as $d(i)$.

Considering the whole, an obvious equation is: $d(1)+\cdots+d(10)=\left|A_{1}\right|+\left|A_{2}\right|+\cdots+\left|A_{k}\right|=5+5+\cdots+5=5 k$. To find the range of $k$, we only need to determine the range of each $d(i)$.

Below, we prove that $d(i) \leqslant 4 (i=1,2, \cdots, 10)$.
In fact, for $i \in M$, consider all pairs $(i, j)$ (where $j \neq i$) and all 5-element subsets containing $i$. On one hand, $i$ can form 9 pairs $(i, j)$ (where $j \neq i$) with other elements in $M$, and each pair $(i, j)$ can appear at most twice in the 5-element subsets $A_{1}, A_{2}, \cdots, A_{k}$. Therefore, all pairs $(i, j)$ can appear at most $2 \times 9 = 18$ times.

On the other hand, there are $d(i)$ subsets containing $i$ in $A_{1}, A_{2}, \cdots, A_{k}$. Since $\left|A_{j}\right|=5 (j=1, 2, \cdots, k)$, each subset containing $i$ has exactly 4 pairs $(i, j)$. Thus, $4 \cdot d(i) \leqslant 18$, so $d(i) \leqslant 4$.

Therefore, $5 k = d(1) + \cdots + d(10) \leqslant 4 \times 10 \Rightarrow k \leqslant 8$.
Finally, when $k=8$, the following 8 5-element sets indeed satisfy the requirements:
$A_{1}=\{1,2,3,4,5\}, A_{2}=\{1,6,7,8,9\}, A_{3}=\{1,3,5,6,8\}$, $A_{4}=\{1,2,4,7,9\}, A_{5}=\{2,3,6,7,10\}, A_{6}=\{3,4,7,8,10\}, A_{7}=\{4,5,8,9,10\}, A_{8}=\{2,5,6,9,10\}$, hence $k_{\min }=8$.

Alternative solution: Examine the element relationship table of the set, and connect every two "1"s in the same column with a line segment.
On one hand, since each column has 5 "1"s, each column connects $\mathrm{C}_{5}^{2}=10$ line segments, so the table connects a total of $10 \cdot k=10 k$ line segments. On the other hand, each line segment corresponds to two rows (the rows where the endpoints are located), and according to the problem, each pair of rows corresponds to at most two line segments, so the number of line segments is no more than $2 \mathrm{C}_{10}^{2}=90$, thus $10 k \leqslant 90$, so $k \leqslant 9$.

If $k=9$, then each pair of rows corresponds to exactly two line segments, i.e., each type of line segment appears exactly twice, so the "1"s in each row appear in pairs, and the number of "1"s in each row is even. Therefore, the total number of "1"s in the table is even. But each column has 5 "1"s, and 9 columns have a total of $9 \cdot 5=45$ (an odd number) "1"s, which is a contradiction, so $k \leqslant 8$.