Consider a polygon of k sides with n points on each side (no point on the vertices)
(i) How many straight lines can be drawn from these ‘kn’ points such that each line passes through exactly 2 of the given points.
(ii) if ‘T’ is the maximum number of triangles that can be drawn from these kn points as vertices then find the value of ‘T’.
(iii) if ‘Q’ is the maximum number of quadrilaterals that can be drawn from these ‘kn’ points as vertices then find the value of Q

(Q). Consider a polygon of k sides with n points on each side (no point on the vertices)
(i) How many straight lines can be drawn from these ‘kn’ points such that each line passes through exactly 2 of the given points.
(ii) if ‘T’ is the maximum number of triangles that can be drawn from these kn points as vertices then find the value of ‘T’.
(iii) if ‘Q’ is the maximum number of quadrilaterals that can be drawn from these ‘kn’ points as vertices then find the value of Q.
Solution:-
(i). From k sides we have to select 2 sides & this can be done in ᵏC₂ ways.
Now from each selected side we have to select 1 point & that can be done in ⁿC₁×ⁿC₁ ways.
So total number of such straight lines

(ii) From k.n points without any restriction we can get maximum ᵏⁿC₃ triangles.
From this we have to reduce the number of triangles that we can not get due to collinear points which is equal to k.ⁿC₃
So total number of such triangle is 
= ᵏⁿC₃ – k.ⁿC₃               Answer

Alternately: In order to draw a triangle we need 3 non-collinear points. So here we have two cases:-
case(i):  If three selected points from three different sides:

case(ii):  if two points are selected from one side while one point is selected from different sides:-

= 2.ᵏC₂(ⁿC₂×ⁿC₁)
So total number of such triangle is 
= ᵏC₃×ⁿC₁×ⁿC₁×ⁿC₁ + 2.ᵏC₂.(ⁿC₂×ⁿC₁)                Answer

(iii). From kn points without any restriction we can get maximum ᵏⁿC₄ quadrilaterals.
From this we have to reduce the number of quadrilaterals that we can not get if three points are collinear & this can happen in two cases:-
case(i):  when all 4 selected points are collinear
 = k × ⁿC₄
case(ii): when three points are collinear
= k × ⁿC₃ ×(k-1) × n
So total number of such quadrilaterals is :

= ᵏⁿC₄ – k.ⁿC₄ – k. ⁿC₃.(k-1).n           Answer

Alternately: In order to draw a quadrilateral we need 4 points such that no three of them are collinear. Here we have three cases:
case(i): if 4 points selected from 4 different sides: 

case(ii): if two points are selected from one side while 1 point is selected from two different sides:

case(iii): if two points are selected from one side while two points are selected from different side:

ᵏC₂×ⁿC₂×ⁿC₂

So total number of such quadrilateral is 
= ᵏC₄×(ⁿC₁×ⁿC₁×ⁿC₁×ⁿC₁) + ᵏC₃×(ⁿC₂×ⁿC₁×ⁿC₁×3) + ᵏC₂×ⁿC₂×ⁿC₂              Answer