7・68 Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ and satisfy
(1) $\left|A_{i}\right|=5, i=1,2, \cdots, k$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 2,1 \leqslant i<j \leqslant k$.

Find the maximum value of $k$.
[Solution 1] The following 6 subsets
$$
\begin{array}{l}
\{1,2,3,4,5\},\{1,2,6,7,8\},\{1,3,6,9,10\}, \\
\{2,4,7,9,10\},\{3,5,7,8,9\},\{4,5,6,8,10\}
\end{array}
$$

satisfy conditions (1) and (2), so the maximum value of $k$ is at least 6.
If there are 7 subsets $A_{1}, A_{2}, \cdots, A_{7}$ that satisfy conditions (1) and (2), then there are 35 elements in these 7 subsets. By the pigeonhole principle, there is an element in $S$ that belongs to at least 4 of the 7 subsets. Without loss of generality, let $1 \in A_{i}, i=1,2,3,4$. At this time, let
$$
A_{i}^{*}=A_{i}-\{1\}, i=1,2,3,4,
$$

then $\left|A_{i}^{*}\right|=4$ and $\left|A_{i}^{*} \cap A_{j}^{*}\right| \leqslant 1,1 \leqslant i<j \leqslant 4$. This indicates that among the 16 elements in $A_{1}^{*}, A_{2}^{*}, A_{3}^{*}, A_{4}^{*}$, there are at most 6 pairs of repeated elements, thus there are at least 10 different elements. Adding 1, there are at least 11 different elements, which is impossible.
In summary, the maximum value of the number of subsets $k$ is 6.
[Solution 2] Suppose there are $k$ subsets that satisfy the conditions (1) and (2) in the problem, and let $i$ belong to $x_{i}$ of these $k$ subsets, $i=1,2, \cdots, 10$. If $i \in A_{j}, i \in A_{k}, j \neq k$, then $i$ is called a repeated pair. Thus, the number of repeated pairs caused by the number $i$ is $C_{x_{i}}^{2}$. The total number of repeated pairs caused by the 10 elements in $S$ is $C_{x_{1}}^{2}+C_{x_{2}}^{2}+\cdots+C_{x_{10}}^{2}, x_{1}+x_{2}+\cdots+x_{10}=5 k$.

On the other hand, there are at most two repeated pairs between any two subsets, so there are at most $2 C_{k}^{2}$ repeated pairs among the $k$ subsets. Therefore, we have
$$
C_{x_{1}}^{2}+C_{x_{2}}^{2}+\cdots+C_{x_{10}}^{2} \leqslant 2 C_{k}^{2} .
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& C_{x_{1}}^{2}+C_{x_{2}}^{2}+\cdots+C_{x_{10}}^{2} \\
= & \frac{1}{2}\left\{x_{1}\left(x_{1}-1\right)+x_{2}\left(x_{2}-1\right)+\cdots+x_{10}\left(x_{10}-1\right)\right\} \\
= & \frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}\right)-\frac{1}{2}\left(x_{1}+x_{2}+\cdots+x_{10}\right) \\
= & \frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}\right)-\frac{5}{2} k \\
\geqslant & \frac{1}{20}(5 k)^{2}-\frac{5}{2} k=\frac{5}{4} k(k-2) .
\end{aligned}
$$

From (1) and (2), we get
$$
\frac{5}{4}(k-2) \leqslant k-1 .
$$

Solving (3) gives $k \leqslant 6$. This indicates that there are at most 6 subsets.
The example at the beginning of Solution 1 shows that there can be 6 subsets that satisfy the requirements of the problem. Therefore, the maximum value of $k$ is 6.