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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)
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B
mn2(m+n2)
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C
mn2(m+n1)
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D
Non of these
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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
$$=\cfrac { m\left( m-1 \right)  }{ 2 } .n+\cfrac { n\left( n+1 \right)  }{ 2 } .m+mn$$
$$=\cfrac { mn }{ 2 } \left[ m-1+n-1 \right] +mn$$
$$=mn\left[ \cfrac { m+n }{ 2 } -1+1 \right] =\cfrac { mn }{ 2 } \left( m+n \right) $$

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