There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. The number of triangles formed by these points is:

Options

A. A. 578
B. B. 576
C. C. 805
D. D. 806

Answer

To solve the problem of finding the number of triangles that can be formed from 18 points in a plane, where no three points are collinear except for five collinear points, we can follow these steps:

### Step 1: Understand the Problem
We have a total of 18 points. Out of these, 5 points are collinear, meaning they lie on the same straight line. To form a triangle, we need to select 3 points, and at least two of these points must not be collinear.

### Step 2: Calculate Total Combinations of Points
First, we calculate the total number of ways to choose 3 points from the 18 points without considering the collinearity restriction. This can be done using the combination formula:

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

For our case, \( n = 18 \) and \( r = 3 \):

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

### Step 3: Subtract Invalid Combinations
Next, we need to subtract the combinations that do not form a triangle, specifically the combinations that can be formed using the 5 collinear points. We cannot form a triangle using 3 collinear points.

The number of ways to choose 3 points from the 5 collinear points is:

\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 4: Calculate Valid Triangles
Now, we subtract the invalid combinations from the total combinations:

\[ \text{Valid triangles} = \text{Total combinations} - \text{Invalid combinations} \]
\[ \text{Valid triangles} = 816 - 10 = 806 \]

### Step 5: Conclusion
Thus, the total number of triangles that can be formed by these points is **806**.