Question

There are $3$ coplanar (non-concurrent) lines $A,B$ and $C$, from these lines $10,12,11$ points are chosen respectively, such that no points other than points lying on the same line are collinear, then the total number of triangles that can be formed using these points are

• A. ${}^{32}{C}_2+{}^{32}{C}_3-({}^{10}{C}_3+{}^{12}{C}_3+{}^{11}{C}_3)$
• B. ${}^{33}{C}_3-{(}^{11}{C}_4+{}^{12}{C}_3)$
• C. ${}^{10}{C}_2\times {}^{23}{C}_1{+}^{11}{C}_2\times {}^{22}{C}_1{+}^{12}{C}_2\times {}^{21}{C}_1{+}^{10}{C}_1\times {}^{11}{C}_1\times {}^{12}{C}_1$
• D. ${}^{33}{C}_3-{(}^{10}{C}_3+{}^{10}{C}_4+{}^{12}{C}_3)$

Answer 1

Answer: (A), (C)

The correct options are
A ${}^{32}{C}_2+{}^{32}{C}_3-({}^{10}{C}_3+{}^{12}{C}_3+{}^{11}{C}_3)$
C ${}^{10}{C}_2\times {}^{23}{C}_1{+}^{11}{C}_2\times {}^{22}{C}_1{+}^{12}{C}_2\times {}^{21}{C}_1{+}^{10}{C}_1\times {}^{11}{C}_1\times {}^{12}{C}_1$

For triangles we need three points $\to$
number of ways of selecting three points $={}^{33}{C}_3$
But out of them $10,12,11$ points lies on the line and hence are collinear
So total triangles will be $={}^{33}{C}_3-({}^{10}{C}_3+{}^{12}{C}_3+{}^{11}{C}_3)$
The above equation can also be expressed as sum of equation$\left(1\right)+\left(2\right)+\left(3\right)+\left(4\right)$
where,
${}^{10}{C}_2\times {}^{23}{C}_1...\left(1\right)$
(two points from line $A$)
${}^{12}{C}_2\times {}^{21}{C}_1...\left(2\right)$
(two points from line $B$)
${}^{11}{C}_2\times {}^{22}{C}_1...\left(3\right)$
​​​​​​​(two points from line $C$)
${}^{10}{C}_1\times {}^{12}{C}_1\times {}^{11}{C}_1...\left(4\right)$
(one point from each line)