Question

There are 5 points \( P_1, P_2, P_3, P_4, P_5 \) on the side \( AB \), excluding \( A \) and \( B \), of a triangle \( \triangle ABC \). Similarly, there are 6 points \( P_6, P_7, \dots, P_{11} \) on the side \( BC \) and 7 points \( P_{12}, P_{13}, \dots, P_{18} \) on the side \( CA \) of the triangle.The number of triangles that can be formed using the points \( P_1, P_2, \dots, P_{18} \) as vertices, is:

• A. 776
• B. 751
• C. 796
• D. 771

Answer 1

The Correct Option is B

Total Points on the Triangle: There are 5 points on \(AB\), 6 points on \(BC\), and 7 points on \(CA\), for a total of:
\[ 5 + 6 + 7 = 18 \text{ points} \] 
Selecting 3 Points to Form a Triangle: To form a triangle, we need to select any 3 points out of these 18 points. The total ways to choose 3 points out of 18 is:
\[ \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \] 
Subtracting Collinear Points: 
We need to subtract cases where the selected 3 points are collinear, as these do not form a triangle: 

Points on \(AB\): There are \(\binom{5}{3} = 10\) ways to select 3 collinear points from the 5 points on \(AB\). 

Points on \(BC\): There are \(\binom{6}{3} = 20\) ways to select 3 collinear points from the 6 points on \(BC\). 

Points on \(CA\): There are \(\binom{7}{3} = 35\) ways to select 3 collinear points from the 7 points on \(CA\). 

Therefore, the number of ways to select collinear points is: \[ 10 + 20 + 35 = 65 \] 

Calculating the Number of Triangles: Subtract the collinear cases from the total selections: \[ 816 - 65 = 751 \]