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

Answer: (B)

751

Answer 2

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: $816 - 65 = 751$