JEE Main 2013 (Online) 23rd AprilQuestion 1
On the sides $AB$, $BC$, $CA$ of a $\triangle ABC$, $3,4,5$ distinct points (excluding vertices $A,B,C$) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are :
Permutation and Combination
Solution
Concept. Three points make a triangle unless all three lie on the same side.Formulas used. (1) ${}^nC_3=\frac{n(n-1)(n-2)}{6}$Step 1 - Points. $3+4+5=12$. Step 2 - All triples. ${}^{12}C_3=220$. Step 3 - Collinear. ${}^3C_3+{}^4C_3+{}^5C_3=1+4+10=15$. Step 4 - Subtract. $220-15=205$. Answer: B