Question

If the sides $AB$, $BC$ and $CA$ of a $△ABC$ have $a$, $b$ and $c$ points lying on them respectively, then the number of triangles that can be constructed using these points as vertices, if

• A. two vertices lie on same side $={}^{a+b+c}{C}_3-({}^a{C}_3+{}^b{C}_3+{}^c{C}_3+abc)$
• B. two vertices lie on the same side $={}^{a+b+c}{C}_3-abc$
• C. all the vertices lie on different sides $=abc$
• D. all the vertices lie on different sides $={}^{a+b+c}{C}_3-({}^a{C}_3+{}^b{C}_3+{}^c{C}_3)$

Answer 1

Answer: (A), (C)

The correct options are
A two vertices lie on same side $={}^{a+b+c}{C}_3-({}^a{C}_3+{}^b{C}_3+{}^c{C}_3+abc)$
C all the vertices lie on different sides $=abc$

Total number of triangles $=$ total number of ways of choosing $3$ points $-$ number of ways of choosing all the $3$ points from $AB$ or $BC$ or $CA$
$={}^{a+b+c}{C}_3-({}^a{C}_3+{}^b{C}_3+{}^c{C}_3)$
Number of triangle formed such that all the vertices lies on different sides $={}^a{C}_1\cdot {}^b{C}_1\cdot {}^c{C}_1=abc$

$\therefore$Number of triangles formed such that two of the vertices lies on same side $={}^{a+b+c}{C}_3-({}^a{C}_3+{}^b{C}_3+{}^c{C}_3+abc)$