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Two lines intersect at O. Points $A_1,A_2,........A_n$ are taken on one of them and $B_1,B_2,........B_n$ on other. The no. of triangles that can be drawn from with the help of these 2n points and point 'O' is ?

$\begin{array}{1 1} ^{2n+1}C_3 \\ n^3+n^2+2 \\ ^{2n+1}C_3-2.^nC_3+1 \\ n^3 \end{array}$

Can you answer this question?
 
 

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To form a $\Delta$  Case 1:
we take any 2 points out of n+1 points $A_1,A_2,......A_n, and \:O$ and
one point out of the $n$ points $B_1,B_2,........B_n$
or
case 2:
we take any 2 points out of n+1 points $B_1,B_2,......B_n$ and $O$ and
one point out of the $n$ points $A_1,A_2,........A_n$
 
$\therefore$ These $\Delta^s$ are $ ^{n+1}C_2\times n$ in number each case.
 
This includes those triangles,which are formed with one vertex at O and one
point on each line.These triangles are $n\times n=n^2$ in number.
i.e. These $n^2\:\Delta^s$ are counted twice. once with  case 1 and once again with case 2.
 
$\therefore$  The required no. of $\Delta^s$ = $ ^{n+1}C_2 \times n+^{n+1}C_2\times n\:-\:n^2$
$=2.n.^{n+1}C_2-n^2$
$=2.n.\large\frac{(n+1)n}{2}-n^2$
$=n^3+n^2-n^2=n^3$

 

answered Aug 27, 2013 by rvidyagovindarajan_1
edited Aug 31, 2013 by rvidyagovindarajan_1
 

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