Question

Two lines intersect at $O$. Points $A_i$ and $B_i(i=1,2,....,n)$ are taken on these two lines respectively, the number of triangles that can be drawn with the help of these $2n+1$ points is

• A. $n$
• B. $n^2$
• C. $n^3$
• D. None of these

Answer 1

Answer: (C)

$n^3$

Number of triangles $=$ Taking any two points in $B_1,B_2,...,B_n$ and one point from $A_1,A_2,....,A_n$ or taking any two points from $A_1,A_2,A_3,....A_n$ and one point from $B_1,B_2,...,B_n$ or one point $O$ and one point from $A_1,A_2....A_n$ and one point from $B_1,B_2,...,B_n$
$\therefore$ Required ways
${=}^n{C}_2{\times }^n{C}_1{+}^n{C}_1{\cdot }^n{C}_2{+}^n{C}_1{\times }^n{C}_1\times 1$
$=\frac{n^2(n-1)2}{+}\frac{n^2(n-1)}{2}+n^2$
$=\left(n^2\right(n-1)+n^2(n-1)+2n^2)/2$
$=n^2[2n-2+2]/2=n^3$