Question

The number of triangles that can be formed by $5$ points in one line and $3$ points on another parallel line is

• A. ${}^8{C}_3$
• B. ${}^8{C}_3-{}^5{C}_3$
• C. ${}^8{C}_3-{}^5{C}_3-1$
• D. ${}^8{C}_3-{}^5{C}_3+1$

Answer 1

The correct option is C ${}^8{C}_3-{}^5{C}_3-1$

Total triangles that can be formed from $8$ points $={}^8{C}_3$
Since, total points are $8$, but $5$ are collinear and other $3$ are also collinear.
$\therefore$ Required number of ways $={}^8{C}_3-{}^5{C}_3-{}^3{C}_3$