Question

Let \[{{T}_{n}}\] denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If \[{{T}_{n+1}}-{{T}_{n}}=36\], then n is equal to: -
(a) 2
(b) 5
(c) 6
(d) 8
(e) 9

Answer 1

Hint: Select any three points or vertices to form a triangle. For \[{{T}_{n+1}}\], assume that there are (n + 1) numbers of vertices and apply the formula for selecting 3 vertices out of (n + 1) vertices using combinations. Similarly, for \[{{T}_{n}}\], assume that there are ‘n’ number of vertices and apply the same formula for selecting 3 vertices out of ‘n’ vertices. Use the expression: - \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\] to calculate the value of n.

Complete step by step answer:
Here, we have been provided with the expression, \[{{T}_{n+1}}-{{T}_{n}}=36\], where \[{{T}_{n}}\] denotes the number of triangles which can be formed by using the vertices of a regular polygon of ‘n’ sides.
Now, we know that to draw a triangle we need 3 non – collinear points. Non – collinear points means that the points must not lie on the same line. So, we have to select 3 vertices to draw the required triangle.
Here, \[{{T}_{n+1}}\] denotes the number of triangles which can be formed by using the vertices of a regular polygon of (n + 1) sides. Therefore, applying the formula of combinations to select 3 points from (n + 1) points, we get,
\[\Rightarrow {{T}_{n+1}}={}^{n+1}{{C}_{3}}\] - (1)
Now, \[{{T}_{n}}\] denotes the number of triangles formed by using the vertices of a regular polygon of ‘n’ sides. So, applying the formula of combinations, we get,
\[\Rightarrow {{T}_{n}}={}^{n}{{C}_{3}}\] - (2)
Subtracting equation (2) from (1), we get,
\[\Rightarrow {{T}_{n+1}}-{{T}_{n}}={}^{n+1}{{C}_{3}}-{}^{n}{{C}_{3}}\]
Now, substituting the value of, \[{{T}_{n+1}}-{{T}_{n}}=36\], we get,
\[\Rightarrow {}^{n+1}{{C}_{3}}-{}^{n}{{C}_{3}}=36\]
Using the conversion, \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\], we get,
\[\begin{align}
  & \Rightarrow \dfrac{\left( n+1 \right)!}{\left( n+1-3 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=36 \\
 & \Rightarrow \dfrac{\left( n+1 \right)!}{\left( n-2 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=36 \\
 & \Rightarrow \dfrac{n!}{\left( n-3 \right)!3!}\times \left[ \dfrac{n+1}{n-2}-1 \right]=36 \\
 & \Rightarrow \dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)!}{\left( n-3 \right)!3!}\times \left[ \dfrac{n+1-n+2}{n-2} \right]=36 \\
\end{align}\]
Cancelling the common factors and like terms, we get,
\[\begin{align}
  & \Rightarrow \dfrac{n\times \left( n-1 \right)}{2}=36 \\
 & \Rightarrow {{n}^{2}}-n-72=0 \\
 & \Rightarrow {{n}^{2}}-9n+8n-72=0 \\
 & \Rightarrow \left( n-9 \right)\left( n+8 \right)=0 \\
\end{align}\]
\[\Rightarrow n=9\] or -8
Since, the number of vertices cannot be negative, therefore, n = -8 can be rejected. Therefore, the number of vertices will be 9.
\[\Rightarrow n=9\]

So, the correct answer is “Option e”.

Note: One must note that here we do not have to apply the formula for permutations, that is \[{}^{n}{{P}_{r}}\], because we do not have to arrange 3 points or vertices but we have to select them, so the formula for combination is applied. You must find a way to reduce the calculation by cancelling the common terms otherwise it will be very difficult to solve the equation.