Question

There are 10 points in a plane, of which no three points are colinear expect 4. Then the number of distinct triangles that can be formed by joining any three points of these ten points, such that at least one of the vertices of every triangle formed is from the given 4 colinear points is

• A. 80
• B. 100
• C. 96
• D. 116

Answer 1

The Correct Option is C

Let the 10 points be denoted as $P_1, P_2, \ldots, P_{10}$.

We are given that no three points are collinear except for 4 points, which are collinear. Let the 4 collinear points be $A, B, C, D$.

1. Total Number of Triangles:
The total number of triangles that can be formed by joining any three points of the 10 points is $\binom{10}{3}$. Since 4 points are collinear, we cannot form a triangle using only those 4 points. The number of combinations of choosing 3 points from the 4 collinear points is $\binom{4}{3}$.

The number of triangles that can be formed using any three points from the 10 points is:

$ \binom{10}{3} - \binom{4}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} - \frac{4 \cdot 3 \cdot 2}{3 \cdot 2 \cdot 1} = 120 - 4 = 116 $

2. Triangles with At Least One Vertex from the Collinear Points:
Let $S$ be the set of 10 points, and let $C$ be the set of 4 collinear points. We want to count the number of triangles such that at least one vertex is from $C$.

We can find the total number of triangles that can be formed from the 10 points, and subtract the number of triangles that can be formed from the remaining 6 points (i.e., no vertex from $C$).

Total number of triangles is $\binom{10}{3} - \binom{4}{3} = 120 - 4 = 116$.
Number of triangles that can be formed from the remaining 6 points is $\binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$.

So the number of triangles with at least one vertex from the 4 collinear points is:

$\binom{10}{3} - \binom{6}{3} - \binom{4}{3} = 120 - 20 = 100$

3. Case-by-Case Analysis:
Alternatively, we can consider the cases where the triangle has 1, 2, or 3 vertices from the 4 collinear points.

1 vertex from $C$ and 2 vertices from the other 6 points: $\binom{4}{1} \binom{6}{2} = 4 \cdot \frac{6 \cdot 5}{2} = 4 \cdot 15 = 60$
2 vertices from $C$ and 1 vertex from the other 6 points: $\binom{4}{2} \binom{6}{1} = \frac{4 \cdot 3}{2} \cdot 6 = 6 \cdot 6 = 36$
3 vertices from $C$ and 0 vertices from the other 6 points: $\binom{4}{3} \binom{6}{0} = 4 \cdot 1 = 4$. However, since the 4 points are collinear, we cannot form a triangle from these 4 points. So we exclude this case.

Thus, the total number of triangles with at least one vertex from the 4 collinear points is $60 + 36 = 96$.

Final Answer:
The final answer is $\boxed{96}$.