2・43 Let $p$ be an odd prime. Consider the subsets $A$ of the set $\{1,2, \cdots, 2 p\}$ that satisfy the following two conditions:
（1） $A$ has exactly $p$ elements;
（2） The sum of all elements in $A$ is divisible by $p$.
Find the number of all such subsets $A$.
[Solution]
$$
\begin{aligned}
\text { Let } U & =\{1,2, \cdots, p\}, \\
V & =\{p+1, \cdots, 2 p\}, \\
W & =\{1,2, \cdots \cdots, 2 p\} .
\end{aligned}
$$

If $S, T$ are two $p$-element subsets of $W$, and they are different from $U, V$, and satisfy the following two conditions:
(i) $S \cap V=T \cap V$;
(ii) By appropriate numbering, the elements $S_{1}, S_{2}, \cdots, S_{m}$ of $S \cap U$ and the elements $t_{1}, t_{2}, \cdots, t_{m}$ of $T \cap U$, for some $k \in\{0,1, \cdots, p-1\}$, satisfy the congruence
$$
S_{i}-t_{i} \equiv k(\bmod p) \quad i=1,2, \cdots, m .
$$

Then we agree that the two subsets $S$ and $T$ belong to the same class.
For different subsets $S$ and $T$ in the same class, it is clear that $K \neq 0$, hence
$$
\sum_{i=1}^{p} S_{i}-\sum_{i=1}^{p} t_{i} \equiv m k \not 0(\bmod p) .
$$

For different subsets in the same class, the sum of their elements modulo $p$ has different remainders, so each class contains exactly $p$ subsets, among which only one subset satisfies condition (2).

In summary, among the $C_{2 p}^{p}$ $p$-element subsets of $W=\{1,2, \cdots, 2 p\}$, except for $U$ and $V$, every $p$ subsets are grouped into one class, and each class has exactly 1 subset that satisfies condition (2). Therefore, the total number of subsets of $W$ that satisfy conditions (1) and (2) is
$$
\frac{1}{p}\left(C_{2 p}^{p}-2\right)+2 .
$$