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Question: The number of triangles that can be formed by using the vertices of a regular polygon of \(\left( n+...

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

Explanation

Solution

We first describe how triangles are formed and how we can make triangles by joining the vertices of a nn- sided polygon. Then we will find the formula of choosing rr things out of nn things. If Tn{{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 nn if the number of triangles that can be formed by using the vertices of a regular polygon of (n+3)\left( n+3 \right) sides is 220. A nn- sided polygon has nn vertices and nn 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 nn vertices are non-linear. As we are trying to create triangles by joining the vertices of a nn- sided polygon, we have to choose 3 points out of those (n+3)\left( n+3 \right) points. So,
The number of triangles = n+3C3=220{}^{n+3}{{C}_{3}}=220
Therefore, (n+3)!3!.n!=220\dfrac{\left( n+3 \right)!}{3!.n!}=220
So, it means,
(n+1)(n+2)(n+3)=1320 (n+1)(n+2)(n+3)=12×10×11 (n+1)(n+2)(n+3)=(9+1)(9+2)(9+3) \begin{aligned} & \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{aligned}
By the comparison of both sides it can be written that n=9n=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 nC2n{}^{n}{{C}_{2}}-n.