Let $X$ be a set with $n\ge 2$ elements. Define $\mathcal{P}(X)$ to be the set of all subsets of $X$. Find the number of functions $f:\mathcal{P}(X)\mapsto \mathcal{P}(X)$ such that
$$|f(A)\cap f(B)|=|A\cap B|$$
whenever $A$ and $B$ are two distinct subsets of $X$.

[i] (Sergiu Novac)[/i]
1. **Define the problem and notation:**
   Let \( X \) be a set with \( n \ge 2 \) elements. Define \( \mathcal{P}(X) \) to be the power set of \( X \), i.e., the set of all subsets of \( X \). We need to find the number of functions \( f: \mathcal{P}(X) \to \mathcal{P}(X) \) such that 
   \[
   |f(A) \cap f(B)| = |A \cap B|
   \]
   for any two distinct subsets \( A \) and \( B \) of \( X \).

2. **Establish the properties of \( f \):**
   Note that for any subset \( A \subseteq X \), the function \( f \) must preserve the size of the intersection. This implies that \( |f(A)| = |A| \) for all subsets \( A \subseteq X \). 

3. **Define the map \( \sigma \):**
   Let \( \sigma \) be the map defined by \( f(\{i\}) = \{\sigma(i)\} \) for each \( i \in X \). Since \( f \) preserves the size of the set, \( \sigma \) is well-defined. 

4. **Show that \( \sigma \) is a bijection:**
   To show that \( \sigma \) is a bijection, consider two distinct elements \( i, j \in X \). Since \( f(\{i\}) \cap f(\{j\}) = \emptyset \) for \( i \neq j \), it follows that \( \sigma(i) \neq \sigma(j) \). Therefore, \( \sigma \) is injective. Since \( X \) has \( n \) elements and \( \sigma \) maps each element of \( X \) to a unique element of \( X \), \( \sigma \) is also surjective. Hence, \( \sigma \) is a bijection.

5. **Determine the form of \( f \):**
   For any subset \( A \subseteq X \), if \( i \in A \), then \( |f(A) \cap \{\sigma(i)\}| = |A \cap \{i\}| = 1 \), so \( \sigma(i) \in f(A) \). Conversely, if \( i \notin A \), then \( |f(A) \cap \{\sigma(i)\}| = 0 \), so \( \sigma(i) \notin f(A) \). Therefore, \( f(A) = \{\sigma(i) : i \in A\} \).

6. **Count the number of such functions:**
   Each function \( f \) is uniquely determined by the bijection \( \sigma \). The number of bijections from a set of \( n \) elements to itself is \( n! \) (the number of permutations of \( n \) elements).

Thus, the number of functions \( f: \mathcal{P}(X) \to \mathcal{P}(X) \) satisfying the given condition is \( n! \).

The final answer is \( \boxed{n!} \).