Solution to a) part :
We know for a straight line to form = 25C2
But 7 points are collinear ..Hence there will be duplicacy in 25C2 lines..So we need to avoid the redndancy due to these 7 points and hence we subtract the lines formed by these 7 collinear points..But we need to add 1 at last because these 7 collinear points will lead to the formation of one line..[Collinear points mean points lying on the same line]
Hence number of such lines = 25C2 - 7C2 + 1.
= 280
Solution to B) part :
For a triangle to form we need 3 points.So number of ways it can be done among 25 points = 25C3
But 7 points are collinear and hence wont contribute to a triangle if all of the three points are chosen from these 7 vertices as they form a line not a triangle..Hence we need to subtract these cases.
Hence required number of triangles = 25C3 - 7C3
= 2265
Solution to C) part :
For a quadrilateral to form we need 4 points.So number of ways it can be done among 25 points = 25C4
But 7 points are collinear and hence wont contribute to a quadrilateral if all of the four points are chosen from these 7 vertices as they form a line not a quadrilateral..Hence we need to subtract these cases.
Further there is a possibility that we choose three points out of 7 collinear point and one point out of 18 non collinear points and hence joining these 4 points will result to a traingle and not a quadrilateral..Hence we need to subtract these triangles..
Number of such triangles = 7C3 ( 3 points selected from collinear points ) * 18C1
Hence required number of quadrilaterals = 25C4 - 7C4 - 7C3 * 18C1
= 11985