Let $l_1$ and $l_2$ be two lines intersecting at $P$. If $A_1, B_1, C_1$ are points on $l_1$, and $A_2, B_2,…
Let $l_1$ and $l_2$ be two lines intersecting at $P$. If $A_1, B_1, C_1$ are points on $l_1$, and $A_2, B_2, C_2, D_2, E_2$ are points on $l_2$ and if none of these coincides with $P$, then the number of triangles formed by these eight points, is :
56
55
46
45
Solution
If triangle is formed including point ' $P$ ' the other points must be one from $l_1$ and other point from $l_2$. Number of triangles formed with $P$,
$n\left(E_1\right)={ }^3 C_1 \times{ }^5 C_1=15 \text { ways }$
When $P$ is not included, Number of triangles formed
$n\left(E_2\right)={ }^3 C_2 \times{ }^5 C_1+{ }^3 C_1 \times{ }^5 C_2$
$=15+15=30$
Total number of triangle $=n\left(E_1\right)+n\left(E_2\right)$ $=15+30=45$