[Example 4.5.5] 16 students participate in a math competition, where all the questions are multiple-choice with four options each. After the exam, it is found that any two students have at most one answer in common. How many questions are there at most? Explain your reasoning.
Let there be $k$ questions, and let the four options be $1,2,3,4$.
Suppose $a_{i}$ students answer a certain question with $i, i=1,2,3,4$.
We study $a_{1}, a_{2}, a_{3}, a_{4}$ as a whole.
For a certain question $S$, $a_{1}$ people answer $1, a_{2}$ people answer $2, a_{3}$ people answer $3, a_{4}$ people answer $4$, then
$$
a_{1}+a_{2}+a_{3}+a_{4}=16 .
$$

Now define "pair of two people" as two people who give the same answer to a question.
Consider the total number of "pairs of two people" $A$ who give the same answer to question $S$.
$$
A=C_{a_{1}}^{2}+C_{a_{2}}^{2}+C_{a_{3}}^{2}+C_{a_{4}}^{2}=\frac{1}{2} \sum_{i=1}^{4}\left(a_{i}^{2}-a_{i}\right) .
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\sum_{i=1}^{4} a_{i}^{2} \geqslant \frac{1}{4}\left(\sum_{i=1}^{4} a_{i}\right)^{2}=\frac{1}{4} \times 16^{2}=64 \\
A=\frac{1}{2}\left(\sum_{i=1}^{4} a_{i}^{2}-\sum_{i=1}^{4} a_{i}\right)=\frac{1}{2}\left(\sum_{i=1}^{4} a_{i}^{2}-16\right) \\
\geqslant \frac{1}{2}(64-16)=24
\end{array}
$$

Now consider the total number of "pairs of two people" $B$ for $k$ questions.
Since $A \geqslant 24$, then $B \geqslant 24 k$.
On the other hand, by the problem's condition, any two students' answers can be the same for at most one question, meaning any two people can form at most one "pair of two people", so
$$
\begin{array}{c}
24 k \leqslant B \leqslant C_{16}^{2}=120 . \\
k \leqslant 5 .
\end{array}
$$

From the above, we estimate that there are at most 5 questions. Below, we construct a set of answers for 5 questions that meet the problem's requirements.
Let $P_{i}$ represent the $i$-th student, $i=1,2, \cdots, 16$.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Question No. & $P_{1}$ & $P_{2}$ & $P_{3}$ & $P_{4}$ & $P_{5}$ & $P_{6}$ & $P_{7}$ & $P_{8}$ & $P_{9}$ & $P_{10}$ & $P_{11}$ & $P_{12}$ & $P_{13}$ & $P_{14}$ & $P_{15}$ & $P_{16}$ \\
\hline One & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & 4 & 4 & 4 & 4 \\
\hline Two & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\
\hline Three & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 \\
\hline Four & 1 & 2 & 3 & 4 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 & 2 & 1 & 4 & 3 \\
\hline Five & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 \\
\hline
\end{tabular}

So there are at most 5 questions.
This problem is solved by estimating the total number of "pairs of two people" for a single question $S$ and for $k$ questions as a whole, leading to the conclusion that $k \leqslant 5$.