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 ABC. Similarly there are 6 points `P_6, P_7,….,P_(11)` on the side BC and 7 points `P_(12), P_(13),…, P_(18)` on the side CA of the triangle. The number of triangles, that can be formed using the points `P_1, P_2,…., P_(18)` as vertices, is:

Options

A. 776
B. 771
C. 751
D. 796

Answer

To find the number of triangles that can be formed using the points \( P_1, P_2, \ldots, P_{18} \) as vertices, we will follow these steps:

### Step 1: Calculate the total number of combinations of points
The total number of points is 18. The number of ways to choose 3 points from these 18 points is given by the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

So, we calculate:

\[ \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \]

### Step 2: Subtract the combinations that do not form triangles
We need to subtract the combinations of points that lie on the same line, as they cannot form a triangle.

1. **Points on side AB**: There are 5 points \( P_1, P_2, P_3, P_4, P_5 \). The number of ways to choose 3 points from these 5 is:

\[ \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \]

2. **Points on side BC**: There are 6 points \( P_6, P_7, P_8, P_9, P_{10}, P_{11} \). The number of ways to choose 3 points from these 6 is:

\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

3. **Points on side CA**: There are 7 points \( P_{12}, P_{13}, P_{14}, P_{15}, P_{16}, P_{17}, P_{18} \). The number of ways to choose 3 points from these 7 is:

\[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 3: Total combinations that do not form triangles
Now, we sum these combinations:

\[ \text{Total non-triangle combinations} = \binom{5}{3} + \binom{6}{3} + \binom{7}{3} = 10 + 20 + 35 = 65 \]

### Step 4: Calculate the final number of triangles
Now, we subtract the non-triangle combinations from the total combinations:

\[ \text{Number of triangles} = \binom{18}{3} - (\binom{5}{3} + \binom{6}{3} + \binom{7}{3}) = 816 - 65 = 751 \]

Thus, the number of triangles that can be formed using the points \( P_1, P_2, \ldots, P_{18} \) as vertices is **751**.