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: \[ 816 - 65 = 751 \]

Answer 2

Step 1: Understand the problem.
We are given 18 points in total:
- 5 points \( P_1, P_2, P_3, P_4, P_5 \) on side \( AB \),
- 6 points \( P_6, P_7, \dots, P_{11} \) on side \( BC \),
- 7 points \( P_{12}, P_{13}, \dots, P_{18} \) on side \( CA \).
We need to find the number of triangles that can be formed using these points as vertices.

Step 2: Conditions for forming a triangle.
To form a triangle, the three vertices must not be collinear. Since all the points are on the sides of the triangle \( \triangle ABC \), any three points chosen from these 18 points will be collinear if they lie on the same side.

Step 3: Total number of ways to choose 3 points from 18 points.
The total number of ways to select 3 points from 18 points is given by the combination formula:
\[ \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \]
Step 4: Subtract the number of degenerate triangles (collinear points).
We need to subtract the number of ways to select 3 collinear points (i.e., points on the same side).
- From side \( AB \), we can select 3 points from the 5 points in \( \binom{5}{3} = 10 \) ways.
- From side \( BC \), we can select 3 points from the 6 points in \( \binom{6}{3} = 20 \) ways.
- From side \( CA \), we can select 3 points from the 7 points in \( \binom{7}{3} = 35 \) ways.
Thus, the total number of degenerate (collinear) triangles is:
\[ 10 + 20 + 35 = 65 \]
Step 5: Calculate the number of valid triangles.
The total number of valid triangles is:
\[ \binom{18}{3} - 65 = 816 - 65 = 751 \]

Final Answer:
The number of triangles that can be formed is:
\[ \boxed{751} \]