Question

There are m points on a straight line AB & n points on the line AC none of them being the point A. Triangles are formed with these points as vertices, when (i) A is excluded (ii) A is included.

The ratio of number of triangles in the two cases is?

• A. $\frac{m+n-2}{m+n}$
• B. $\frac{m+n-2}{m+n-1}$
• C. $\frac{m+n-2}{m+n+2}$
• D. $\frac{m(n-1)}{(m+1)(n+1)}$

Answer 1

The correct option is A $\frac{m+n-2}{m+n}$

Consider triangle without vertex

we can choose $2$ vertices from line $AB$ and one vertex from $A$ the possibilities are

${}^m{C}_2\times n$

We can choose $2$ vertices from line $AC$ and one vertex from $AB$ the possibilities are:

${}^n{C}_2\times m$

As anyone of the above can be done so number of possibilities is

${}^m{C}_2\times n+{}^n{C}_2\times m$

Solving

${}^m{C}_2\times {}^n{C}_2\times m$

$=\frac{m!}{2!(m-2)!}\times n+\frac{n!}{2!(n-2)!}\times m$

$=\frac{m(m-1)}{2}\times n+\frac{n(n-1)}{2}\times m$

$=\frac{mn(m+n-2)}{2}$

Consider triangles with vertex $A$

As one vertex is $A$, we can choose one vertex from $AC$ and one from $AB$ the possibilities are

$l\times m\times n$

$=mn$

Number of triangle is mn(m+n)/2$

Taking the ratio of $1$ and $2$

$\frac{mn(n+m-2)}{2}/\frac{mn(m+n)}{2}$

$\frac{m+n-2}{m+n}$