Question

The number of triangles that can be formed by using the vertices of a regular polygon of $\left( n+3 \right)$ sides is 220. Then $n$ is equal to:
1. 8
2. 9
3. 10
4. 11
5. 12

Answer 1

Hint: We first describe how triangles are formed and how we can make triangles by joining the vertices of a $n$- sided polygon. Then we will find the formula of choosing $r$ things out of $n$ things. If ${{T}_{n}}$ be the number of triangles made, then we will use the formula to find the number of ways 3 points can be chosen from 10 points. We will put the values in the equation and solve it to find the solution of the problem.

Complete step-by-step solution:
According to the question it is asked to find $n$ if the number of triangles that can be formed by using the vertices of a regular polygon of $\left( n+3 \right)$ sides is 220. A $n$- sided polygon has $n$ vertices and $n$ sides. We know that in order to form a triangle we need three points which are non-linear. We also have that no three points out of those $n$ vertices are non-linear. As we are trying to create triangles by joining the vertices of a $n$- sided polygon, we have to choose 3 points out of those $\left( n+3 \right)$ points. So,
The number of triangles = ${}^{n+3}{{C}_{3}}=220$
Therefore, $\dfrac{\left( n+3 \right)!}{3!.n!}=220$
So, it means,
$\begin{align}
  & \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)=1320 \\
 & \Rightarrow \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)=12\times 10\times 11 \\
 & \Rightarrow \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)=\left( 9+1 \right)\left( 9+2 \right)\left( 9+3 \right) \\
\end{align}$
By the comparison of both sides it can be written that $n=9$.
Hence the correct answer is option 2.

Note: We have to remember that in the case of triangles, we don’t have to worry, but if we are asked about diagonal, where we need to find any 2 points, then at the end, we have to subtract the number of sides from the option as they will be considered as diagonals. So, the answer will be ${}^{n}{{C}_{2}}-n$.