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(A) Case $I$: When $A$ is excluded.Number of triangles = (selection of $2$ points from $AB$ and $1$ point from $AC$) + (selection of $1$ point from $A

Vedclass · Accepted Answer

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

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(A) Case $I$: When $A$ is excluded.Number of triangles = (selection of $2$ points from $AB$ and $1$ point from $AC$) + (selection of $1$ point from $AB$ and $2$ points from $AC$).Number of triangles = $^mC_2 	imes ^nC_1 + ^mC_1 	imes ^nC_2 = \frac{m(m-1)}{2} 	imes n + m 	imes \frac{n(n-1)}{2} = \frac{mn}{2}(m-1+n-1) = \frac{mn(m+n-2)}{2}$.Case $II$: When $A$ is included.Triangles can be formed by taking $A$ as one vertex and two other points from the lines $AB$ and $AC$.If we take $A$ as a vertex,we need one point from $AB$ and one point from $AC$ to form a triangle.Number of triangles = $^mC_1 	imes ^nC_1 = mn$.Total triangles when $A$ is included = (Triangles not involving $A$) + (Triangles involving $A$) = $\frac{mn(m+n-2)}{2} + mn = \frac{mn}{2}(m+n-2+2) = \frac{mn(m+n)}{2}$.Required ratio = $\frac{	ext{Case } I}{	ext{Case } II} = \frac{\frac{mn(m+n-2)}{2}}{\frac{mn(m+n)}{2}} = \frac{m+n-2}{m+n}$.

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

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