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There are 25 points on a plane of which 7 are collinear.

Find

a. How many straight lines can be formed?

b. How many triangles can be formed?

c. How many quadrilaterals can be formed?

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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  =   25C -  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   7C18C1

                                                                    =     11985

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