Question

From $25$ points on a plane, $8$ are on a straight line, $7$ are on another straight line and $10$ are on a third straight line.
Then the number of triangles that can be drawn by connecting some three points from these $25$ is

• A. ${}^{25}{C}_3$
• B. ${}^{25}{C}_3-({}^8{C}_3+{}^7{C}_3+{}^{10}{C}_3)$
• C. ${}^{25}{C}_3+({}^8{C}_3+{}^7{C}_3+{}^{10}{C}_3)$
• D. ${}^8{C}_3$+${}^7{C}_3+{}^{10}{C}_3$

Answer 1

The correct option is B ${}^{25}{C}_3-({}^8{C}_3+{}^7{C}_3+{}^{10}{C}_3)$

From the $25$ points on a plane we can form ${}^{25}{C}_3$ triangles but it is given that $8$ are on one straight line, $7$ are on another straight line and $10$ more are on another straight line, so we can't form triangles on joining points which are on straight lines.
So, we have to subtract $({}^8{C}_3+{}^7{C}_3+{}^{10}{C}_3)$ from the total.

Hence, we can form ${}^{25}{C}_3-({}^8{C}_3+{}^7{C}_3+{}^{10}{C}_3)$ triangles.