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 how many points of intersection the line AB has with the circles drawn through the points A, B, C, D, and E, excluding the points A and B themselves.

### Step-by-Step Solution:

1. **Identify the Points and Circles**:
We have five points: A, B, C, D, and E. Since no three points are collinear and no four points are concyclic, we can form circles using any three of these points.

2. **Calculate the Number of Circles**:
The number of ways to choose 3 points from 5 points can be calculated using the combination formula:
\[ \text{Number of circles} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Thus, there are 10 different circles that can be formed using the points A, B, C, D, and E.

3. **Determine Intersections with Line AB**:
Each circle will intersect the line AB at two points. However, since points A and B are already on the line, we need to exclude these from our count of intersections.

4. **Count Unique Intersection Points**:
For each circle formed by choosing points A, C, D, for example, the line AB will intersect this circle at two points. One of these points is A, and the other point will be a new point (let's call it P1). Therefore, for each circle, we get one unique intersection point apart from A and B.

5. **Calculate Total Unique Intersection Points**:
Since there are 10 circles, and each circle contributes 1 unique intersection point (excluding A and B), the total number of unique intersection points is:
\[ \text{Total unique intersection points} = 10 \]

6. **Final Count**:
Thus, the total number of points of intersection on the line AB, apart from A and B, is:
\[ \text{Total} = 10 \]

### Conclusion:
The number of points of intersection on the line AB, apart from A and B, is **10**.