There are 12 points in a plane of which 5 are collinear. Except these five points no three are collinear, then

Options

A. The number of quadrilaterals one can form using these points is 420
B. the number of Deltas one can form using these points is 210
C. the number of lines one can form using these points is 57
D. the number of lines one can form using these points is 56

Explanation

To solve the problem, we need to find the number of quadrilaterals, triangles, and lines that can be formed using the given points in the plane, considering that 5 of these points are collinear.

### Step 1: Calculate the number of quadrilaterals
A quadrilateral can be formed by selecting 4 points. However, we cannot select all 4 points from the 5 collinear points, as they do not form a quadrilateral.

1. **Total points**: 12
2. **Collinear points**: 5

The number of ways to choose 4 points from 12 is given by \( \binom{12}{4} \).

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]

Now, we need to subtract the cases where all 4 points are chosen from the 5 collinear points:

\[ \binom{5}{4} = 5 \]

Thus, the number of valid quadrilaterals is:

\[ \text{Valid Quadrilaterals} = \binom{12}{4} - \binom{5}{4} = 495 - 5 = 490 \]

### Step 2: Calculate the number of triangles
A triangle can be formed by selecting 3 points. We need to subtract the cases where all 3 points are chosen from the 5 collinear points.

1. **Total ways to choose 3 points from 12**:

\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

2. **Subtract the cases where all 3 points are collinear**:

\[ \binom{5}{3} = 10 \]

Thus, the number of valid triangles is:

\[ \text{Valid Triangles} = \binom{12}{3} - \binom{5}{3} = 220 - 10 = 210 \]

### Step 3: Calculate the number of lines
A line can be formed by selecting 2 points. We need to subtract the cases where both points are chosen from the 5 collinear points.

1. **Total ways to choose 2 points from 12**:

\[ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \]

2. **Subtract the cases where both points are collinear**:

\[ \binom{5}{2} = 10 \]

Thus, the number of valid lines is:

\[ \text{Valid Lines} = \binom{12}{2} - \binom{5}{2} = 66 - 10 = 56 \]

### Final Results
- Number of valid quadrilaterals: **490**
- Number of valid triangles: **210**
- Number of valid lines: **56**