There are 16 points in a plane of which 6 points are collinear and no other 3 points are collinear.Then the number of quadrilaterals that can be formed by joining these points is

Answer

To solve the problem of finding the number of quadrilaterals that can be formed by joining 16 points in a plane, where 6 points are collinear and no other 3 points are collinear, we can break it down into several cases based on how many of the collinear points are included in the quadrilateral.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 16 points in total, out of which 6 are collinear. A quadrilateral cannot be formed if it has 3 or more collinear points. Therefore, we will consider three cases:
- Case 1: 0 collinear points
- Case 2: 1 collinear point
- Case 3: 2 collinear points

2. **Case 1: 0 Collinear Points**:
- We need to select 4 points from the 10 non-collinear points.
- The number of ways to choose 4 points from 10 is given by the combination formula \( \binom{n}{r} \).
- Thus, the number of quadrilaterals in this case is:
\[ \text{Case 1} = \binom{10}{4} \]

3. **Case 2: 1 Collinear Point**:
- We choose 1 point from the 6 collinear points and 3 points from the 10 non-collinear points.
- The number of ways to choose 1 from 6 is \( \binom{6}{1} \) and the number of ways to choose 3 from 10 is \( \binom{10}{3} \).
- Thus, the number of quadrilaterals in this case is:
\[ \text{Case 2} = \binom{6}{1} \times \binom{10}{3} \]

4. **Case 3: 2 Collinear Points**:
- We choose 2 points from the 6 collinear points and 2 points from the 10 non-collinear points.
- The number of ways to choose 2 from 6 is \( \binom{6}{2} \) and the number of ways to choose 2 from 10 is \( \binom{10}{2} \).
- Thus, the number of quadrilaterals in this case is:
\[ \text{Case 3} = \binom{6}{2} \times \binom{10}{2} \]

5. **Calculating Each Case**:
- Calculate \( \binom{10}{4} \):
\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
- Calculate \( \binom{6}{1} \times \binom{10}{3} \):
\[ \binom{6}{1} = 6, \quad \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
\[ \text{Case 2} = 6 \times 120 = 720 \]
- Calculate \( \binom{6}{2} \times \binom{10}{2} \):
\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15, \quad \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
\[ \text{Case 3} = 15 \times 45 = 675 \]

6. **Total Number of Quadrilaterals**:
- Now, add the results from all three cases:
\[ \text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 210 + 720 + 675 = 1605 \]

### Final Answer:
The total number of quadrilaterals that can be formed is **1605**.