Question

There are m points on a straight line AB and n points on another line AC, none of them being the point A. Triangles are formed from these points as vertices when (i) A is excluded (ii) A is included. Then the ratio of the number of triangles in the two cases is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Case I: When A is excluded.

Number of triangles = selection of 2 points from AB and one point from AC + selection of one point from AB and two points from AC

=${}^m{C}_2{}^n{C}_1+{}^m{C}_1{}^n{C}_2$ = $\frac{1}{2}(m+n-2)mn$ ..........(i)

Case II: When A is included.

The triangles with one vertex at A = selection of one point from AB and one point from AC = mn.

$\therefore$ Number of triangles = $mn+\frac{1}{2}mn(m+n-2)$ = $\frac{1}{2}mn(m+n)$ ................(ii)

$\therefore$ Required ratio = $\frac{(m+n-2)}{m+n}$