Example 2.5.2 Find the generating function for the following combination problems, and solve the corresponding problems:
(1) Find the number of $k$-combinations of $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$;
(2) Find the number of $k$-combinations of $\left\{\infty \cdot a_{1}, \infty \cdot a_{2}, \cdots, \infty \cdot a_{n}\right\}$;
(3) Find the number of 10-combinations of $\{3 \cdot a, 4 \cdot b, 5 \cdot c\}$;

where $\{3 \cdot a, 4 \cdot b, 5 \cdot c\}$ represents a set composed of 3 $a$'s, 4 $b$'s, and 5 $c$'s, etc.
(1) This is a combination problem of ordinary sets.
(2) This is a $k$-combination problem with repetition of $n$. Let the $i$-th element among $n$ elements be repeated $x_{i}$ times $(i=1,2, \cdots, n)$, then the $k$-combination problem with repetition of $n$ elements can be transformed into the problem of finding the number of non-negative integer solutions to $x_{1}+x_{2}+\cdots+x_{n}=k$.
(3) In $\{3 \cdot a, 4 \cdot b, 5 \cdot c\}$, let $a$ be taken $x$ times, $b$ be taken $y$ times, and $c$ be taken $z$ times, then the 10-combination problem can be transformed into the problem of finding the number of solutions to $x+y+z=10,0 \leqslant x \leqslant 3,0 \leqslant y \leqslant 4,0 \leqslant z \leqslant 5$.
Using the generating function method, the above three types of combination problems can be uniformly processed. First, find the generating function.
(1) $f(x)=(1+x)^{n}$, the expansion of which, each $x^{k}$ is $x^{m_{1}} x^{m_{2}} \cdots x^{m_{n}}=x^{k}, m_{i}$ $=0,1$.
(2) $f(x)=\left(1+x+x^{2}+\cdots\right)^{n}=\frac{1}{(1-x)^{n}}$,
its expansion, each $x^{k}$ is $x^{m_{1}} x^{m_{2}} \cdots x^{m_{n}}=x^{k}$, $m_{1}+m_{2}+\cdots+m_{n}=k \quad\left(m_{i} \geqslant 0\right)$.
(3)
$$
\begin{aligned}
f(x)= & \left(1+x+x^{2}+x^{3}\right)\left(1+x+x^{2}+x^{3}+x^{4}\right) \\
& \left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right) \\
= & \frac{1-x^{4}}{1-x} \frac{1-x^{5}}{1-x} \frac{1-x^{6}}{1-x} .
\end{aligned}
$$

Then expand to find the coefficient $f_{k}$.
(1) $(1+x)^{n}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k} x^{k},$ the coefficient of $x^{k}$ is $\mathrm{C}_{n}^{k}$.
(2) $f(x)=\frac{1}{(1-x)^{n}}$ to find the coefficient $f_{k}$ of $x^{k}$ after expansion, it can be done as follows (using calculus):
$$
\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{k}+\cdots,
$$

Differentiate both sides: $\frac{1}{(1-x)^{2}}=1+2 x+\cdots+k x^{k-1}+(k+1) x^{k}+\cdots$,
$$
\begin{array}{l} 
\frac{2}{(1-x)^{3}}= 2+\cdots+k(k-1) x^{k-2}+(k+1) k x^{k-1} \\
+(k+2)(k+1) x^{k}+\cdots \\
\cdots \cdots \\
\frac{(n-1)!}{(1-x)^{n}}=(n-1)!+\cdots+(k+n-1) \\
(k+n-2) \cdots(k+2)(k+1) x^{k}+\cdots .
\end{array}
$$

So $\frac{1}{(1-x)^{n}}=\sum_{k=0}^{\infty} \mathrm{C}_{k+n-1}^{n-1} x^{k}$,
$$
f_{k}=\mathrm{C}_{k+n-1}^{n-1} \text {. }
$$
(3) $f(x)=\left(1-x^{4}\right)\left(1-x^{5}\right)\left(1-x^{6}\right) /(1-x)^{3}$
$$
=\left(1-x^{4}-x^{5}-x^{6}+x^{9}+x^{10}+x^{11}-x^{15}\right)\left(1+\sum_{k=1}^{\infty} C_{k+3-1}^{3-1} x^{k}\right)
$$
The coefficient of $x^{10}$, $f_{10}=\mathrm{C}_{12}^{2}-\mathrm{C}_{8}^{2}-\mathrm{C}_{7}^{2}-\mathrm{C}_{6}^{2}+\mathrm{C}_{3}^{2}+1=6$.