Consider a rectangle ABCD having 5, 7, 6, 9 points in the integer of the line segments AB, CD, BC, DA respectively. Let “α” be the number of triangles having these points from different sides as vertices and “β” be  the number of quadrilaterals having these points from different sides, as vertices. Then β−α is equal to:

Given that ABCD is a rectangle as shown belowThe number of triangles is equal to the number of ways of choosing three points, each from one side  and the number of quadrilaterals is equal to the number of ways of selecting four vertices from four different sides Number of triangles is =C5,1⋅C7,1⋅C6,1+C7,1⋅C6,1⋅C9,1                                     +C6,1⋅C9,1⋅C5,1+C5,1⋅C7,1⋅C9,1=5⋅7⋅6+7⋅6⋅9+6⋅9⋅5+5⋅7⋅9=210+378+270+315=1173Number of Quadrilaterals is =C5,1⋅C7,1⋅C6,1⋅C9,1=5⋅7⋅6⋅9=1890Therefore the required number is 717