We are given a triangle $ABC$ with points on its sides. The side $AB$ has 6 points, side $BC$ has 7 points, and side $AC$ has 8 points. None of these points coincide with the vertices of the triangle. We need to calculate how many triangles can be formed with one vertex on side $AB$, one vertex on side $BC$, and one vertex on side $AC$.

Step 1: Understand the problem setup

We are asked to form triangles such that:

• One vertex of the triangle lies on side $AB$,
• One vertex lies on side $BC$,
• One vertex lies on side $AC$.

The points on each side are distinct and we are selecting exactly one point from each side to form a triangle.

Step 2: Count the number of triangles

To form a triangle, we simply choose one point from side $AB$, one point from side $BC$, and one point from side $AC$.

• We can choose 1 point from the 6 points on side $AB$,
• We can choose 1 point from the 7 points on side $BC$,
• We can choose 1 point from the 8 points on side $AC$.

The number of ways to form a triangle by selecting one point from each side is simply the product of the number of choices from each side. Therefore, the total number of triangles is:

$\text{Number of triangles} = 6 \times 7 \times 8 = 336$

Final Answer:

There are $336$ triangles that can be formed with one vertex on $AB$, one vertex on $BC$, and one vertex on $AC$.

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Do you have any questions or would you like further details?

Here are 5 related questions for further exploration:

1. What if the points on one of the sides were doubled? How would this affect the total number of triangles?
2. How many triangles can be formed if two vertices must lie on the same side of the triangle?
3. If the points on side $AB$ were reduced to 3, how many triangles could be formed?
4. What if one of the sides has an unknown number of points? How would you express the total number of triangles?
5. How would the problem change if we allowed points to coincide with the vertices of the triangle?

Tip: Always break down complex counting problems by focusing on each independent choice separately, then use multiplication to combine them.