Example 6 There are 16 students taking an exam, all questions are multiple-choice, each with 4 options. After the exam, it was found that: any two students have at most one question with the same answer. How many questions are there at most? (33rd


Analysis and Solution: Let there be $n$ questions. We prove that $n_{\text {max }}=5$.
For each question, the answers of 16 students form a sequence of length 16, containing only $1, 2, 3, 4$ (answer codes). Arrange the sequences corresponding to $n$ questions into an $n \times 16$ number table.

Notice the condition "any two people have at most one question with the same answer." We can calculate the pairs of identical answers in each question (each row). For convenience in describing the problem, for each row, if two answer codes are the same, we connect these two numbers with a line segment, called a same-color line segment. Suppose a certain row has $x$ 1s, $y$ 2s, $z$ 3s, and $t$ 4s, where $x+y+z+t=16$, then the row has $\mathrm{C}_{x}^{2}+\mathrm{C}_{y}^{2}+\mathrm{C}_{z}^{2}+\mathrm{C}_{t}^{2} \geqslant \mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{2}=24$ (segments) of same-color line segments. Thus, $n$ rows have at least $24n$ same-color line segments. According to the problem, any two same-color line segments in the $n$ rows have different projections, so $24n \leqslant \mathrm{C}_{16}^{2}=120$, hence $n \leqslant 5$.

If $n=5$, then all the inequalities hold as equalities, so the number table that meets the conditions should satisfy the following two points: $\square$
(1) Each row has 4 1s, 4 2s, 4 3s, and 4 4s;
(2) Different same-color line segments do not overlap in the $n$ rows.

Assume the first row is $1111 \quad 2222 \quad 3333 \quad 4444$ (considered as 4 groups), then each group in the other rows is a permutation of 1234 (because same-color line segments do not overlap). A natural permutation is 1234, and we can fill in as many 1234s as possible in the table. It can be found that the corresponding groups in the second row can all be filled with 1234, and the first group (the first 4 columns) can all be filled with 1234 except for the first row. Continuing this way, we can obtain the following number table that meets the conditions, which shows that $n=5$ is possible.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline \begin{tabular}{c} 
Question Number
\end{tabular} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\
\hline 1 & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & 4 & 4 & 4 & 4 \\
\hline 2 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\
\hline 3 & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 \\
\hline 4 & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 \\
\hline 5 & 1 & 2 & 3 & 4 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 & 4 & 3 & 2 & 1 \\
\hline
\end{tabular}