Question

Let ${{T}_{n}}$ be the number of all possible triangles formed by joining vertices of an $n$ -sided regular polygon. If ${{T}_{n+1}}-{{T}_{n}}=19$ then the value of $n$ is
A. $7$
B. $5$
C. $10$
D. $8$

Answer 1

Hint: In this problem we need to calculate the value of $n$ for the given conditions. We have given that the ${{T}_{n}}$ is the number of all possible triangles formed by joining the vertices of $n$ sided polygon. We know that to form a triangle we need to join $3$ vertices. So we will calculate the value of ${{T}_{n}}$ by calculating the value of selecting $3$ vertices from $n$ vertices. After having the value ${{T}_{n}}$ we can calculate the value of ${{T}_{n+1}}$ and substitute both the values in the given equation and simplify them to get the required result.

Complete step by step solution:
Given that ${{T}_{n}}$ is the number of all possible triangles formed by joining the vertices of $n$ sided polygon.
We know that a $n$sided polygon has $n$ vertices and to form a triangle by joining the vertices of the polygon we need to join any $3$ of the vertices. So the number of possible triangle that can be formed in the $n$ sided polygon is given by
${{T}_{n}}={}^{n}{{C}_{3}}$
From the above equation, the value of ${{T}_{n+1}}$ will be
${{T}_{n+1}}={}^{n+1}{{C}_{3}}$
Substituting the values of ${{T}_{n}}$ and ${{T}_{n+1}}$ in the given equation which is ${{T}_{n+1}}-{{T}_{n}}=19$ , then we will get
${}^{n+1}{{C}_{3}}-{}^{n}{{C}_{3}}=10$
Using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ in the above equation, then we will have
$\begin{align}
  & \dfrac{\left( n+1 \right)!}{\left( n+1-3 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=10 \\
 & \Rightarrow \dfrac{\left( n+1 \right)!}{\left( n-2 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=10 \\
\end{align}$
Writing the value $\left( n+1 \right)!$ as $\left( n+1 \right)\times \left( n+1-1 \right)!=\left( n+1 \right)n!$ and $\left( n-2 \right)!$ as $\left( n-2 \right)\times \left( n-2-1 \right)!=\left( n-2 \right)\left( n-3 \right)!$ in the above equation, then we will get
$\dfrac{\left( n+1 \right)n!}{\left( n-2 \right)\left( n-3 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=10$
Taking $\dfrac{n!}{\left( n-3 \right)!3!}$ as common from the two terms in LHS, and then we will have
$\begin{align}
  & \dfrac{n!}{\left( n-3 \right)!3!}\left( \dfrac{n+1}{n-2}-1 \right)=10 \\
 & \Rightarrow \dfrac{n!}{\left( n-3 \right)!3!}\left( \dfrac{n+1-n+2}{n-2} \right)=10 \\
 & \Rightarrow \dfrac{n!}{\left( n-3 \right)!3!}\left( \dfrac{3}{n-2} \right)=10 \\
\end{align}$
Writing the value $3!$ as $3\times \left( 3-1 \right)!=3\times 2!$ in the above equation, then we will get
$\begin{align}
  & \dfrac{n!}{\left( n-3 \right)!3\times 2!}\left( \dfrac{3}{n-2} \right)=10 \\
 & \Rightarrow \dfrac{n!}{\left( n-2 \right)\left( n-3 \right)!2!}=10 \\
\end{align}$
We have the value $\left( n-2 \right)\left( n-3 \right)!$ as $\left( n-2 \right)!$ and using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ in the above equation, then we will have
$\begin{align}
  & \dfrac{n!}{\left( n-2 \right)!2!}=10 \\
 & \Rightarrow {}^{n}{{C}_{2}}=10 \\
\end{align}$
We can write the value $10$ as ${}^{5}{{C}_{2}}$ in the above equation, and then we will have
${}^{n}{{C}_{2}}={}^{5}{{C}_{2}}$
Equating on both sides of the above equation, then we will get
$\therefore n=5$
Hence option – B is the correct answer.

Note: In this problem we have solved the equation ${}^{n+1}{{C}_{3}}-{}^{n}{{C}_{3}}=10$ by using the formulas of permutation and combination. We have a direct formula for this equation which is ${}^{n+1}{{C}_{r}}-{}^{n}{{C}_{r}}={}^{n}{{C}_{r-1}}$ . we can use this formula to simplify the equation and to get the required result.