In a plane, there are 11 points, out of which 5 are collinear. Find the number of triangles made by these points.

Answer

To solve the problem of finding the number of triangles that can be formed from 11 points in a plane, where 5 of those points are collinear, we can follow these steps:

### Step 1: Calculate the total number of triangles that can be formed from 11 points.
To find the total number of triangles that can be formed from 11 points, we use the combination formula \( nCk \), which gives us the number of ways to choose \( k \) items from \( n \) items without regard to the order of selection.

For triangles, we need to choose 3 points from the 11 points:
\[ \text{Total triangles} = \binom{11}{3} \]

### Step 2: Calculate the number of triangles that cannot be formed due to collinearity.
Since 5 of the points are collinear, any triangle formed using these 5 points will not be valid. We need to calculate the number of ways to choose 3 points from these 5 collinear points:
\[ \text{Collinear triangles} = \binom{5}{3} \]

### Step 3: Subtract the invalid triangles from the total triangles.
The total number of valid triangles can be found by subtracting the number of collinear triangles from the total triangles:
\[ \text{Valid triangles} = \binom{11}{3} - \binom{5}{3} \]

### Step 4: Calculate the values of the combinations.
Now, we will compute the values of \( \binom{11}{3} \) and \( \binom{5}{3} \):

1. Calculate \( \binom{11}{3} \):
\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

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

### Step 5: Find the total number of valid triangles.
Now, substitute the values back into the equation for valid triangles:
\[ \text{Valid triangles} = 165 - 10 = 155 \]

Thus, the total number of triangles that can be formed from the given points is **155**.

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