Question

Number of triangles formed by joining the vertices of $n$ sided polygon which has no side in common with that of the polygon.

• A. $\frac{n(n-3)}{2}$
• B. $\frac{(n-4)(n-5)}{3!}$
• C. $\frac{n(n-4)(n-5)}{3!}$
• D. None of these

Answer 1

Answer: (C)

$\frac{n(n-4)(n-5)}{3!}$

For the case we consider the equation
Total number of triangles formed
$=$ Triangle having no side common$+$ triangle having exactly one side common$+$ triangles having exactly two sides common $($with those of the polygon$)$
$\therefore$ Number of triangles having no sides common with that of polygon
$=($Total Number of triangles i.e ${}^n{C}_3)-$Number of $\delta$ exactly one side common $-$ Number of triangles having exactly two sides common.
Now Number of $\Delta$ having exactly one side common $=n(n-4)$
and Number of triangles having exactly two sides common.
For this we have to select three consecutive vertices of the polygon,

If vertices of polygon are marked by $(A_1,A_2.....A_n)$
i.e$A_1{A}_2{A}_3,A_1{A}_2{A}_3{A}_4......A_n{A}_1{A}_2$

which can be done by $n$ ways.
$\therefore$ Required triangles having no sides common with the of the polygon
${=}^n{C}_3-n(n-4)-n$
$=\frac{n}{6}[n^2-9n+20]=\frac{n(n-4)(n-5)}{3!}$