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
mn2(m+n)
B
mn2(m+n2)
C
mn2(m+n1)
D
Non of these

Solution

## The correct option is C $$\dfrac {mn}{2}(m+n)$$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$$=\cfrac { m\left( m-1 \right) }{ 2 } .n+\cfrac { n\left( n+1 \right) }{ 2 } .m+mn$$$$=\cfrac { mn }{ 2 } \left[ m-1+n-1 \right] +mn$$$$=mn\left[ \cfrac { m+n }{ 2 } -1+1 \right] =\cfrac { mn }{ 2 } \left( m+n \right)$$Maths

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