Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, without regard to order, is given by the combination formula: $^nC_r = \frac{n!}{r!(n-r)!}$.
• Triangle Formation: Three points form a triangle if and only if they are non-collinear.
• Principle of Inclusion-Exclusion (Geometric Combinations): The number of triangles formed by a set of points is the total number of ways to choose three points minus the number of ways to choose three collinear points.

Step-by-Step Solution

1. Understand the Given Information and Identify Total Points

We are given a triangle ABC with 3 interior points on side AB, 5 interior points on side BC, and 6 interior points on side CA. "Interior points" means the points lie strictly between the vertices of the triangle. The vertices A, B, and C are not included in these counts. We need to find the total number of triangles that can be formed using these points as vertices. First, we determine the total number of distinct points.

Total number of points ($N$) = $3 + 5 + 6 = 14$

2. Calculate the Total Number of Ways to Choose 3 Points

We need to choose 3 points out of the 14 distinct points. This can be done in $^{14}C_3$ ways. This is the total number of possible combinations of 3 points without considering whether they are collinear.

${}^{14}{C_3} = \frac{14!}{3!(14-3)!} = \frac{14!}{3!11!} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 14 \times 13 \times 2 = 364$

3. Identify and Calculate Invalid Combinations (Collinear Points)

Since three collinear points cannot form a triangle, we need to subtract the number of combinations of three collinear points from the total number of combinations. The points on each side of the original triangle are collinear.

• Invalid combinations from side AB: There are 3 points on side AB. The number of ways to choose 3 points from these 3 collinear points is $^3C_3$. ${}^{3}{C_3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1$

• Invalid combinations from side BC: There are 5 points on side BC. The number of ways to choose 3 points from these 5 collinear points is $^5C_3$. ${}^{5}{C_3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

• Invalid combinations from side CA: There are 6 points on side CA. The number of ways to choose 3 points from these 6 collinear points is $^6C_3$. ${}^{6}{C_3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

4. Calculate the Total Number of Valid Triangles

The total number of triangles is obtained by subtracting the total number of invalid combinations from the total number of ways to choose 3 points.

Total number of invalid combinations = $^3C_3 + ^5C_3 + ^6C_3 = 1 + 10 + 20 = 31$.

Number of triangles = (Total ways to choose 3 points) - (Total invalid combinations) Number of triangles = $^{14}C_3 - (^3C_3 + ^5C_3 + ^6C_3)$ Number of triangles = $364 - 31$ Number of triangles = $333$

Common Mistakes & Tips

• Misinterpreting "Interior Points": Ensure you understand that "interior points" usually excludes the vertices of the triangle. Including the vertices will lead to a significantly different calculation.
• Forgetting Collinearity: The most common mistake is to forget to subtract the combinations of collinear points. Always identify and subtract these invalid combinations.
• Combination Formula: Ensure you are comfortable with the combination formula and its calculation.

Summary

To find the number of triangles, we calculated the total number of ways to choose 3 points from the 14 available points and then subtracted the number of ways to choose 3 collinear points on each side of the original triangle. This gives us the total number of valid triangles that can be formed.

The final answer is $\boxed{333}$, which corresponds to option (C).