The problem asks how many triangles can be formed from points placed on the sides of triangle $\Delta ABC$, where:

• There are 6 points on side $AB$,
• 7 points on side $BC$,
• 8 points on side $AC$.

None of the points coincide with the vertices of the triangle.

Step-by-step breakdown:

1. Total Points: The total number of points on the sides of $\Delta ABC$ is: $6 \,(\text{points on } AB) + 7 \,(\text{points on } BC) + 8 \,(\text{points on } AC) = 21 \, \text{points}.$

2. Selecting 3 Points: We need to select 3 points to form a triangle. However, these 3 points must not be collinear (on the same line).

   • The number of ways to choose 3 points from 21 is given by the combination formula: $\binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330.$

3. Collinear Points (Invalid Triangles):

   • A triangle cannot be formed if all 3 points lie on the same side of the triangle. So, we subtract the cases where all 3 points are on the same side.

   • Points on $AB$: The number of ways to choose 3 points from the 6 points on $AB$ is: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.$

   • Points on $BC$: The number of ways to choose 3 points from the 7 points on $BC$ is: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.$

   • Points on $AC$: The number of ways to choose 3 points from the 8 points on $AC$ is: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.$

4. Valid Triangles: To get the total number of triangles, we subtract the cases where the points are collinear (on the same side of the triangle) from the total number of ways to choose 3 points. $\text{Valid triangles} = 1330 - (20 + 35 + 56) = 1330 - 111 = 1219.$

Final Answer:

There are 1219 valid triangles that can be formed.

Would you like further details or explanations on any of the steps?

Here are 5 related questions to further explore this topic:

1. What is the general formula for calculating the number of triangles formed from points on the sides of a triangle?
2. How does the number of collinear points affect the total count of triangles?
3. If one side had no points, how would that impact the solution?
4. How would the result change if the points could coincide with the vertices of the triangle?
5. Can the solution method be generalized to polygons with more sides?

Tip: When dealing with combinatorics, always remember to subtract the "invalid" or "forbidden" cases after calculating the total possibilities!