Let $X$ be a set containing $n$ elements. Find the number of ordered triples $(A,B, C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$.
1. **Count the number of ordered triples $(A, B, C)$ such that $A \subseteq B \subseteq C$:**

   For each element $x \in X$, there are 4 possibilities:
   - $x$ belongs to all three sets $A$, $B$, and $C$.
   - $x$ belongs to $B$ and $C$ but not to $A$.
   - $x$ belongs to $C$ only.
   - $x$ belongs to none of the sets.

   Since there are $n$ elements in $X$, the total number of such triples is:
   \[
   4^n
   \]

2. **Count the number of ordered triples $(A, B, C)$ such that $A \subseteq B = C$:**

   For each element $x \in X$, there are 3 possibilities:
   - $x$ belongs to both $A$ and $B = C$.
   - $x$ belongs to $B = C$ but not to $A$.
   - $x$ belongs to none of the sets.

   Since there are $n$ elements in $X$, the total number of such triples is:
   \[
   3^n
   \]

3. **Subtract the number of triples where $B = C$ from the total number of triples:**

   The number of ordered triples $(A, B, C)$ such that $A \subseteq B \subseteq C$ and $B \neq C$ is:
   \[
   4^n - 3^n
   \]

The final answer is $\boxed{4^n - 3^n}$