Question 5 Let $p$ be an odd prime. Find the number of subsets $A$ of the set $\{1,2$, $\cdots, 2 p\}$ that satisfy the following conditions:
(1) $A$ has exactly $p$ elements;
(2) The sum of all elements in $A$ is divisible by $p$.
Let $\sigma(A)$ denote the sum of all elements of $A$, and $[mp]$ denote $\{1,2, \cdots, mp\}$. Then the number of subsets $A$ of $[mp]$ such that $|A|=p$ and $p \mid \sigma(A)$ is
$$
C_{m}=\frac{1}{p}\left[\mathrm{C}_{mp}^{p}+m(p-1)\right].
$$

In fact, consider the polynomial
$$
F(x, y)=(1+x y)\left(1+x y^{2}\right) \cdots\left(1+x y^{np}\right).
$$

Each term in the product contributing to $x^{k} y^{\sigma}$ corresponds to a subset (possibly empty) $A=\left\{i_{1}, i_{2}, \cdots, i_{k}\right\} \subseteq[mp]$, where $|A|=k$ and $\sigma(A)=\sigma$.

Let $\left[x^{p} y^{\sigma}\right] F(x, y)$ denote the coefficient of $x^{p} y^{\sigma}$ in $F(x, y)$. Therefore,
$$
C_{m}=\sum_{p i \sigma}\left[x^{p} y^{\sigma}\right] F(x, y).
$$

Let $\varepsilon=\mathrm{e} \frac{2 \pi \mathrm{i}}{p}$. By the multisection formula, we have
$$
\begin{array}{l}
C_{m}=\left[x^{p}\right] \frac{1}{p} \sum_{j=0}^{p-1} F\left(x, \varepsilon^{j}\right) \\
=\left[x^{p}\right] \frac{1}{p} \sum_{j=0}^{p-1} \prod_{k=1}^{mp}\left(1+x \varepsilon^{kj}\right) \\
=\left[x^{p}\right] \frac{1}{p} \sum_{j=0}^{p-1} \prod_{k=0}^{p-1}\left(1+x \varepsilon^{kj}\right)^{m} \\
=\left[x^{p}\right] \frac{1}{p}\left[(1+x)^{mp}+(p-1)\left(1+x^{p}\right)^{m}\right] \\
=\frac{1}{p}\left[\mathrm{C}_{mp}^{p}+m(p-1)\right].
\end{array}
$$

Let $m=2$, and we get the answer to the original problem:
$$
C_{2}=\frac{1}{p}\left[\mathrm{C}_{2p}^{p}+2(p-1)\right].
$$