Given are $n$ $(n > 2)$ points on the plane such that no three of them are collinear. In how many ways this set of points can be divided into two non-empty subsets with non-intersecting convex envelops?
To solve the problem of determining the number of ways to divide \( n \) points on a plane (with no three points collinear) into two non-empty subsets with non-intersecting convex hulls, we can proceed as follows:

1. **Understanding the Problem**:
   - We need to divide \( n \) points into two non-empty subsets such that their convex hulls do not intersect.
   - This is equivalent to finding a line that separates the points into two groups.

2. **Defining the Function**:
   - Let \( f(n) \) be the number of ways to divide \( n \) points into two subsets with non-intersecting convex hulls.

3. **Base Case**:
   - For \( n = 2 \), there is only one way to divide the points into two subsets, each containing one point. Thus, \( f(2) = 1 \).

4. **Inductive Step**:
   - Consider adding a new point \( A_{n+1} \) to the set of \( n \) points \( A_1, A_2, \ldots, A_n \).
   - We need to count the number of ways to divide the \( n+1 \) points into two subsets with non-intersecting convex hulls.

5. **Case Analysis**:
   - **Case 1**: \( A_{n+1} \) and \( A_n \) are in the same subset.
     - The number of ways to divide the remaining \( n \) points is \( f(n) \).
   - **Case 2**: \( A_{n+1} \) and \( A_n \) are in different subsets.
     - We can choose any of the \( n \) points to be in the same subset as \( A_{n+1} \), and the remaining points will be in the other subset.
     - This gives us \( n \) ways to divide the points.

6. **Recurrence Relation**:
   - Combining the two cases, we get:
     \[
     f(n+1) = f(n) + n
     \]

7. **Solving the Recurrence**:
   - We start with \( f(2) = 1 \).
   - Using the recurrence relation, we can compute \( f(n) \) for larger \( n \):
     \[
     \begin{align*}
     f(3) &= f(2) + 2 = 1 + 2 = 3, \\
     f(4) &= f(3) + 3 = 3 + 3 = 6, \\
     f(5) &= f(4) + 4 = 6 + 4 = 10, \\
     &\vdots \\
     f(n) &= 1 + 2 + 3 + \cdots + (n-1).
     \end{align*}
     \]

8. **Summation Formula**:
   - The sum of the first \( n-1 \) positive integers is given by:
     \[
     1 + 2 + 3 + \cdots + (n-1) = \frac{(n-1)n}{2}
     \]
   - Therefore, we have:
     \[
     f(n) = \binom{n}{2}
     \]

Conclusion:
The number of ways to divide \( n \) points into two non-empty subsets with non-intersecting convex hulls is given by \( \boxed{\binom{n}{2}} \).