Example 6 There are 10 sets of test papers, each set containing 4 questions, and at most one question is the same between any two sets. Among these test papers, what is the minimum number of different questions?
(2005, Taiwan Mathematical Olympiad)
Consider each test paper as a vertex, these vertices form a set $X$; consider each question as a vertex, these vertices form a set $Y$.

If a test paper $x \in X$ contains a question $y \in Y$, connect a line between the corresponding vertices $x$ and $y$. This results in a bipartite graph $X+Y$.

Let $Y$ have $n$ vertices, with degrees $x_{1}, x_{2}, \cdots, x_{n}$.

Since each test paper has 4 questions, each vertex in $X$ has a degree of 4. Therefore,
$x_{1}+x_{2}+\cdots+x_{n}=4 \times 10=40$.
Furthermore, since at most one question is the same between any two test papers, for any two vertices in $X$, there is at most one vertex in $Y$ that forms an angle with them, meaning the number of such angles is less than or equal to $\mathrm{C}_{10}^{2}$.

In $Y$, the number of angles at the $i$-th vertex is $\mathrm{C}_{x_{i}}^{2}$, so the total number of angles is
$\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2}$.
Thus, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant \mathrm{C}_{10}^{2}$.
Combining this with the previous equation, we can simplify to
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leqslant 2 \mathrm{C}_{10}^{2}+40=130 \text {. }
$$

Using a common inequality, we get
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leqslant n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
\Rightarrow 40^{2} \leqslant 130 n \Rightarrow n \geqslant 13 .
\end{array}
$$

Using the aforementioned bipartite graph, it is not difficult to construct an example with $n=13$ (take $x_{1}=x_{2}=4, x_{3}=x_{4}=\cdots=x_{12}=3$, $x_{13}=2$).
Therefore, there are at least 13 different questions.
The idea in Example 6 can be used to prove a general conclusion.

Theorem 2 Let a simple graph $G$ have $n$ vertices and $m$ edges. If graph $G$ does not contain $K_{2, r+1}$, then
$$
m \leqslant \frac{n}{4}[1+\sqrt{4 r(n-1)+1}] .
$$

Proof Since graph $G$ does not contain $K_{2, r+1}$, for any two vertices, the number of angles they form is less than or equal to $r$.

On the other hand, let the degrees of the vertices in the graph be $x_{1}$, $x_{2}, \cdots, x_{n}$, then the number of angles in the graph is
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \text {. }
$$

Therefore, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant r \mathrm{C}_{n}^{2}$.
Combining this with $x_{1}+x_{2}+\cdots+x_{n}=2 m$, we get
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=r n(n-1)+2 m \text {. }
$$

As in Example 6, we can derive the inequality
$$
(2 m)^{2} \leqslant n[m(n-1)+2 m] \text {. }
$$

Thus, the theorem is proved.