There are five points A,B,C,D and E. no three points are collinear and no four are concyclic. If the line AB intersects of the circles drawn through the five points. The number of points of intersection on the line apart from A and B is

Explanation

To solve the problem, we need to determine the number of intersection points of the line AB with the circles formed by the points A, B, C, D, and E, excluding the points A and B themselves.

### Step-by-Step Solution:

1. **Identify the Points and Conditions**:
We have five points: A, B, C, D, and E. The conditions state that no three points are collinear and no four points are concyclic.

2. **Determine the Circles**:
Since no four points are concyclic, we can form circles using any three of the five points. The number of ways to choose 3 points from 5 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of points and \( r \) is the number of points to choose.

\[ \text{Number of circles} = \binom{5}{3} = 10 \]

3. **List the Circles**:
The circles that can be formed from the points A, B, C, D, and E are:
- Circle through A, B, C
- Circle through A, B, D
- Circle through A, B, E
- Circle through A, C, D
- Circle through A, C, E
- Circle through A, D, E
- Circle through B, C, D
- Circle through B, C, E
- Circle through B, D, E
- Circle through C, D, E

4. **Intersection of Line AB with Each Circle**:
A line can intersect a circle at most at two points. However, since we are interested in points of intersection other than A and B, we need to consider the intersections of the line AB with each of the circles formed.

- For each circle that does not include points A and B, the line AB will intersect the circle at two points.
- The circles that do not include A and B are:
- Circle ACD
- Circle ACE
- Circle ADE
- Circle BCD
- Circle BCE
- Circle BDE
- Circle CDE

5. **Count the Intersections**:
Each of the circles listed above will contribute 2 intersection points with line AB:
- Circle ACD: 2 points
- Circle ACE: 2 points
- Circle ADE: 2 points
- Circle BCD: 2 points
- Circle BCE: 2 points
- Circle BDE: 2 points
- Circle CDE: 2 points

Thus, the total number of intersection points (excluding A and B) is:
\[ 7 \text{ circles} \times 2 \text{ intersection points per circle} = 14 \text{ intersection points} \]

### Final Answer:
The total number of points of intersection on the line AB, apart from A and B, is **14**.