To solve this problem, let's consider the number of triangles that can be formed without any collinear points first.

Number of triangles with 10 points without any collinear points:
When no three points are collinear, any three points can form a triangle.
So, we have 10 points and we need to choose 3 points to form a triangle.
This can be calculated using the combination formula: C(10, 3) = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.

So, without any collinear points, we can form 120 triangles.

Now, let's consider the case when n points are collinear.

Number of triangles with n collinear points:
When n points are collinear, we cannot form any triangles using these n points.
So, we need to subtract the number of triangles that cannot be formed due to these collinear points from the total number of triangles without any collinear points.

Number of triangles that cannot be formed due to n collinear points:
To find the number of triangles that cannot be formed, we need to choose 3 points from the remaining (10-n) points.
This can be calculated using the combination formula: C(10-n, 3) = (10-n)! / (3! * ((10-n)-3)!) = (10-n)! / (3! * (7-n)!)

Now, the total number of triangles that can be formed with n collinear points is given by:

Total number of triangles = Number of triangles without any collinear points - Number of triangles that cannot be formed due to n collinear points
110 = 120 - (10-n)! / (3! * (7-n)!)

Simplifying this equation, we get:

(10-n)! / (3! * (7-n)!) = 120 - 110
(10-n)! / (3! * (7-n)!) = 10

Now, we can check the possible values of n and find the one that satisfies the equation.

For n = 3, (10-n)! / (3! * (7-n)!) = 7! / (3! * 4!) = 7 * 6 * 5 / (3 * 2 * 1) = 35, which is not equal to 10.

For n = 4, (10-n)! / (3! * (7-n)!) = 6! / (3! * 3!) = 6 * 5 * 4 / (3 * 2 * 1) = 20, which is not equal to 10.

For n = 5, (10-n)! / (3! * (7-n)!) = 5! / (3! * 2!) = 5 * 4 / (2 * 1) = 10, which is equal to 10.

Hence, the value of n is 5. Therefore, the correct answer is option 'C' (5).