Let $m$ and $n$ be two natural numbers. A school has $n$ students eating in the cafeteria. Over the course of the year, the cafeteria has offered 100 different dishes. Each student has tasted 10 different dishes. Furthermore, if we take any two dishes out of the 100, we know that exactly $m$ students have tasted each of the two dishes. Calculate the ratio: $\frac{n}{m}$.
Let $E$ denote the set of students, $M$ the set of menus, and $A$ the set of triplets $(\{a, b\}, e)$ where $a, b \in M, a \neq b$ and $e \in E$ such that $e$ has tasted menu $a$ and menu $b$. Then, if $e_{0} \in E$, we denote $A_{e_{0}}$ as the set of triplets $(\{a, b\}, e) \in A$ such that $e=e_{0}$, and $A_{\left\{a_{0}, b_{0}\right\}}$ as the set of triplets $(\{a, b\}, e) \in A$ such that $\left\{a_{0}, b_{0}\right\}=\{a, b\}$. If $B$ is a set, we denote $|B|$ as the number of elements in $B$. According to the statement, for all $a, b \in M$ such that $a \neq b,\left|A_{\{a, b\}}\right|=m$ and, since each student has tasted 10 menus, for all $e \in E,\left|A_{e}\right|=\binom{10}{2}=45$. Finally, we have:

$$
\begin{aligned}
|A| & =\sum_{e \in E}\left|A_{e}\right| \\
& =n \times 45 \\
|A| & =\sum_{a, b \in M}\left|A_{\{a, b\}}\right| \\
& =\left(\begin{array}{c}
100 \\
2 \\
2
\end{array}\right) m \\
& =4950 m \\
45 n & =4950 m \\
\frac{n}{m} & =110
\end{aligned}
$$