Question

There are m points on a straight line $AB$ and $n$ 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. $\frac{m+n-1}{m+n}$
• C. $\frac{m+n-2}{m+n+2}$
• D. $\frac{m(n-1)}{(m+1)(n+1)}$

Answer 1

The correct option is B $\frac{m+n-1}{m+n}$

Case i) Triangles without vertex A

We must choose 2 vertices from line AB and one from line AC.

or two lines AC and one vertice from line AB.

vertices

${=}^m{C}_2{\times }^n{C}_1{+}^n{C}_2{\times }^m{C}_1$

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

$=\frac{mn}{2}[m-1+n-1]$

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

Case ii) when A is included

(a) we have already 1 vertex (A)

Therefore, we must choose 1 vertex from line AB

and 1 vertex from line AC.

$=1{\times }^m{C}_1{\times }^n{C}_1$

$=mn$

(b) If point A is not selected, then

selection is same as $1^{st}$ case, so total no.

of ways would be $\frac{mn(m+n-2)}{2}$

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

$=mn[1+\frac{m+n-2}{2}]=\frac{mn(m+n)}{2}$

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