The 79th question: Let $\mathrm{p}$ be an odd prime, and consider the set $\{1,2, \ldots, 2 \mathrm{p}\}$ of subsets $A$ that satisfy the following conditions: (1) $A$ contains exactly $\mathrm{p}$ elements; (2) the sum of all elements in $A$ is divisible by $\mathrm{p}$. Try to find the number of all such subsets.
Question 79: Let $p$ be an odd prime, and consider the set $\{1,2, \ldots, 2p\}$. Find the number of subsets $A$ of this set that satisfy the following conditions: (1) $A$ has exactly $p$ elements; (2) The sum of all elements in $A$ is divisible by $p$. (1995 36th IMO Problem)

Solution Method 1: Let $U=\{1,2, \ldots, 2p\}, X=\{1,2, \ldots, p\}, Y=\{p+1, p+2, \ldots, 2p\}$. Then $X \cup Y=U, X \cap Y=\emptyset$, and both $X$ and $Y$ are sets that satisfy the conditions.

For any subset $A$ that satisfies the conditions other than $X$ and $Y$, let $A \cap X=\{x_1, x_2, \ldots, x_t\}, A \cap Y=\{y_1, y_2, \ldots, y_k\}$, then $1 \leq t \leq p-1, 1 \leq k \leq p-1, t+k=p$.

Let $B_i=\{y_1+i, y_2+i, \ldots, y_k+i\} (i=0, 1, 2, \ldots, p-1)$. If $y_s+i > 2p (1 \leq s \leq k)$, then in $B_i$, replace $y_s+i$ with $y_s+i-p$. At this point, $C_i = B_i \cup (A \cap X)$ satisfies: $|C_i|=p$, and the sum of all elements in $C_i$, $S(C_i) \equiv S(A) + ki \pmod{p}$. Since $1 \leq k \leq p-1$, when $i$ runs through a complete residue system modulo $p$, $ki$ also runs through a complete residue system modulo $p$. This means that, given $A \cap X=\{x_1, x_2, \ldots, x_t\}$, the number of $p$-element subsets $A$ that satisfy the conditions is exactly $\frac{1}{p}$ of the number of $p$-element subsets $C$ that satisfy $C \cap X=\{x_1, x_2, \ldots, x_t\}$.

The number of $p$-element subsets of $U$ is $C_{2p}^p$. After removing $X$ and $Y$, the number of subsets is $C_{2p}^p - 2$. Among these, $\frac{1}{p}$ are the $p$-element subsets that satisfy the conditions. Therefore, the total number of subsets that satisfy the conditions is $\frac{1}{p}(C_{2p}^p - 2) + 2$.

Solution Method 2: Let the number of sets $A$ that satisfy the conditions be $M$.
Let $f(x, y) = (1 + xy)(1 + xy^2) \ldots (1 + xy^{2p})$, then $M$ is the sum of the coefficients of all terms of the form $x^p y^{kp} (k \in \mathbb{N})$ in $f(x, y)$.
For a polynomial $g(x)$, let $\langle x^r \rangle g(x)$ denote the coefficient of $x^r$ in $g(x)$.
Let $\omega = e^{\frac{2\pi i}{p}}$, then $\omega^j = \omega^{p+j}, \sum_{j=0}^{p-1} \omega^{tj} = \begin{cases} 0, & \text{if } p \nmid t \\ p, & \text{if } p \mid t \end{cases}$, and $(1 + \omega x)(1 + \omega^2 x) \ldots (1 + \omega^p x) = 1 + x^p$. Thus,
\[
\begin{array}{l}
M = \langle x^p \rangle \frac{1}{p} \left[ \sum_{j=0}^{p-1} f(x, \omega^j) \right] = \langle x^p \rangle \frac{1}{p} \left[ \sum_{j=0}^{p-1} (1 + x \omega^j)(1 + x \omega^{2j}) \ldots (1 + x \omega^{2pj}) \right] \\
= \langle x^p \rangle \frac{1}{p} \left\{ \sum_{j=0}^{p-1} \left[ (1 + x \omega^j)(1 + x \omega^{2j}) \ldots (1 + x \omega^{pj}) \right]^2 \right\} \\
= \langle x^p \rangle \frac{1}{p} \left\{ (1 + x)^{2p} + \sum_{j=1}^{p-1} \left[ (1 + x \omega^j)(1 + x \omega^{2j}) \ldots (1 + x \omega^{pj}) \right]^2 \right\} \\
= \langle x^p \rangle \frac{1}{p} \left\{ (1 + x)^{2p} + \sum_{j=1}^{p-1} \left[ (1 + \omega x)(1 + \omega^2 x) \ldots (1 + \omega^p x) \right]^2 \right\} \\
= \langle x^p \rangle \frac{1}{p} \left[ (1 + x)^{2p} + \sum_{j=1}^{p-1} (1 + x^p)^2 \right] \\
= \langle x^p \rangle \frac{1}{p} \left[ (1 + x)^{2p} + (p-1)(1 + x^p)^2 \right] \\
= \frac{1}{p} \left[ C_{2p}^p + 2(p-1) \right]
\end{array}
\]

Therefore, the number of sets $A$ that satisfy the conditions is $\frac{1}{p} \left[ C_{2p}^p + 2(p-1) \right]$.
Note: According to Method 2, generally, the number of subsets $A$ of $\{1, 2, \ldots, mp\} (m \in \mathbb{Z}^+)$ that satisfy $|A| = p$ and $\sum_{x \in A} x \equiv 0 \pmod{p}$ is $\frac{1}{p} \left[ C_{mp}^p + m(p-1) \right]$.