Question

There are $m$ points on one straight line $AB$ and $n$ points on another straight line $AC$, none of them being $A$. How many triangles can be formed with these point as vertices?

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

Answer 1

Answer: (A)

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

Triangles with no vertex as A

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

Triangles with one vertex at A

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

$\therefore$Total no. of triangles

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

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

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