Question

Let $ABC$ be a triangle. Take $3$ points on $AB,4$ on $BC$ and $5$ on $AC$ such that none of the points are vertices of the triangle $ABC.$ The number of triangles that can be formed using these points is

• A. $60$
• B. $120$
• C. $145$
• D. $205$

Answer 1

The correct option is D $205$

Case $\left(1\right):$ Choosing one point each on $AB,BC,AC$
Number of ways $=3\times 4\times 5=60$

Case $\left(2\right):$ Choosing one point on $AB$ and two on either $BC$ or $AC$
Number of ways $=3\times ({}^4{C}_2+{}^5{C}_2)$
$=3\times (6+10)=48$

Case $\left(3\right):$ Choosing one point on $BC$ and two on either $AC$ or $AB$
Number of ways $=4\times ({}^5{C}_2+{}^3{C}_2)$
$=4\times (10+3)=52$

Case $\left(4\right):$ Choosing one point on $AC$ and two on either $AB$ or $BC$
Number of ways $=5\times ({}^3{C}_2+{}^4{C}_2)$
$=5\times (3+6)=45$

Total number of ways $=60+48+52+45=205$