To solve the problem, we need to find the number of ways to divide 6 points into two groups of 3 such that the triangles formed by the points in each group do not intersect.

1. Selecting Groups: First, we select 3 points from the 6 points. The number of ways to choose 3 points from 6 is given by the binomial coefficient:

   $\binom{6}{3} = \frac{6!}{3! \cdot (6-3)!} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$

2. Avoiding Overcounting: Each selection of 3 points can be paired with the remaining 3 points to form two triangles. However, since the order of the groups does not matter (i.e., grouping $A$ with $B$ is the same as grouping $B$ with $A$), we need to divide by 2 to avoid double counting:

   $\text{Total groups} = \frac{20}{2} = 10$

3. Non-Intersecting Condition: The condition that the triangles formed by the two groups do not intersect is automatically satisfied when no three points are collinear and all points are distinct. This is because the only way two triangles can intersect is if they share a side or vertex, which cannot happen with the given conditions.

Thus, the final count of ways to divide the 6 points into two groups of 3 without the triangles intersecting is:

$\boxed{10}$

Further Questions

1. How does the condition of no three points being collinear affect the triangles?
2. What happens if some of the points are collinear?
3. Can you explain why we divide by 2 when counting the groups?
4. How would the result change if we had 8 points instead?
5. Can you provide a combinatorial proof for this result?

Tip

Always visualize the problem with diagrams, especially in combinatorial geometry, as it helps in understanding configurations and conditions better.