Of the 8 points in a given plane 4 are on circle O and remaining 4 are

The four points on circle O are not collinear; however, the four points on line L are. The three methods to form a quadrilateral, using four of the eight points, are:

1) All four points are from circle O,

2) Three points are from circle O, and one point is from line L, and

3) Two points are from circle O, and two points are from line L.

Let’s now determine the number of ways for each method.

1) All four points are from circle O

The number of ways using this method is 4C4 = 1.

2) Three points from circle O and one point from line L

The number of ways using this method is 4C3 x 4C1 = 4 x 4 = 16.

3) Two points from circle O and two points from line L

The number of ways using this method is 4C2 x 4C2 = 6 x 6 = 36.

Therefore, the total number of quadrilaterals that can be formed is 1 + 16 + 36 = 53.

Alternate Solution:

From a total of 8 points, 4 points can be chosen in 8C4 = 8!/(4!*4!) = 70 ways.

Among these 70 choices, the following will not form a quadrilateral:

1) All the four points are from line L. There’s only one such choice.

2) Three points are from line L and one point from circle O: The three points can be chosen from the line in 4C3 = 4 ways. The one point can be chosen from the circle in 4C1 = 4 ways. Thus, there are 4 x 4 = 16 such choices.

We see that 16 + 1 = 17 of the 70 choices of four points do not form a quadrilateral and 70 - 17 = 53 choices do.

Answer: B