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(Q). There are ‘n’ points in a plane. No three of which are on same straight line. All the possible straight lines are made by joining these ‘n’ points.
(i) What is the maximum number of point of intersection of these straight lines?
(ii) Taking point of intersection of these straight lines as vertices of triangles then what is the maximum number of triangles that can be formed?

Solution:-
(i) Number of straight lines that can be drawn is ⁿC₂ =  \frac{{n(n - 1)}}{2}  = k (say)
Now from k straight lines for maximum number of point of intersection condition is, no two of them are parallel to each other and no three of them are concurrent. So number of point of intersection of these k straight lines is = ᵏC₂ =   \frac{{n(n - 1)}}{2}C₂      Answer

(ii). From above number of point of intersection of these straight lines = \frac{{n(n - 1)}}{2}C₂  = p (say)
Since every straight line is intersected by every other remaining (k-1) straight lines at (k-1) points & these (k-1) points on being in same line are collinear.

total k lines(k-1) point of intersection(k-1) lines

Now number of triangles

C - k.total number of triangletotal number of straight lineNumber of triangle from (k-1) points= - = - Answer