The seventy-fifth question: Let $\mathrm{A}=\{1,2, \ldots, 2012\}, \mathrm{B}=\{1,2, \ldots, 19\}, \mathrm{S}$ be the set of all subsets of $\mathrm{A}$. Find the number of functions $\mathrm{f}: \mathrm{S} \rightarrow \mathrm{B}$ such that for all $\mathrm{A}_{1} 、 A_{2} \in \mathrm{S}$, we have $\mathrm{f}\left(\mathrm{A}_{1} \cap A_{2}\right)=\min \left\{\mathrm{f}\left(\mathrm{A}_{1}\right)\right.$ , $\left.\mathrm{f}\left(\mathrm{A}_{2}\right)\right\}$
The 75th question,
Answer: Let $f(A)=n$, then for any $X \in S, f(X)=f(A \cap X)=\min \{f(A), f(X)\} \leq f(A)=n$. If $X \in S$, and $|X|=2011$, then there are 2012 such $X$. Thus, for such $X$, there are $n^{2012}$ different ways to determine the value of $f(X)$.

We will now prove: When the values of 2012 such $f(X)$ are determined, the values of $f$ on other subsets are uniquely determined. For this, we first prove by induction on $k$: For any $a_{1}, a_{2}, \ldots, a_{k} \in A$, we have
$$
f\left(A \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}\right)=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), f\left(A \backslash\left\{a_{2}\right\}\right), \ldots, f\left(A \backslash\left\{a_{k}\right\}\right)\right\}
$$

When $k=1$, the conclusion is obviously true.
When $k=2$, according to the condition
$$
f\left(A \backslash\left\{a_{1}, a_{2}\right\}\right)=f\left(\left(A \backslash\left\{a_{1}\right\}\right) \cap\left(A \backslash\left\{a_{2}\right\}\right)\right)=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), f\left(A \backslash\left\{a_{2}\right\}\right)\right\}
$$

The conclusion holds.
Assume the conclusion holds for $k \geq 2$, then by the induction hypothesis:
$$
\begin{array}{l}
f\left(A \backslash\left\{a_{1}, a_{2}, \ldots, a_{k+1}\right\}\right) \\
=f\left(A \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\} \cap\left(A \backslash\left\{a_{k+1}\right\}\right)\right) \\
=\min \left\{f\left(A \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}\right), f\left(A \backslash\left\{a_{k+1}\right\}\right)\right\} \\
=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), f\left(A \backslash\left\{a_{2}\right\}\right), \ldots, f\left(A \backslash\left\{a_{k}\right\}\right), f\left(A \backslash\left\{a_{k+1}\right\}\right)\right\}
\end{array}
$$

Thus, the conclusion holds for $k+1$. By mathematical induction, (*) holds.
On the other hand, if (*) holds, then for all $A_{1}, A_{2} \in S$, we have $f(A_{1} \cap A_{2})=\min \{f(A_{1}), f(A_{2})\}$. Since $1 \leq n \leq 19$, the number of functions satisfying the condition is $1^{2012}+2^{2012}+\ldots+19^{2012}$.