# Let ${T_n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If ${T_{n + 1}} - {T_n} = 21$ then n equals

${\text{A}}{\text{. 5}}$

${\text{B}}{\text{. 7}}$

${\text{C}}{\text{. 6}}$

${\text{D}}{\text{. 4}}$

Answer

Verified

359.7k+ views

Hint: - Here we choose three sides from n sides of a polygon by method of selection to form the triangle i.e.${}^n{C_3}$.Then similarly do for the n+1 sides of the polygon. After that apply the condition of the question.

Complete step by step solution:

Given that,

${T_{n + 1}} - {T_n} = 21$

${T_{n + 1}}$ Can be written as ${}^{n + 1}{C_3}$

$ \Rightarrow {}^{(n + 1)}{C_3} - {}^n{C_3} = 21$

$ \Rightarrow {}^n{C_2} + {}^n{C_3} - {}^n{C_3} = 21$

$\because $ We know that ${}^{(n + 1)}{C_r} = {}^n{C_{r - 1}} + {}^n{C_r}$

$ \Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = 21$

$\because $We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$ \Rightarrow \dfrac{{n \times (n - 1) \times (n - 2)!}}{{2 \times (n - 2)!}} = 21$

$

\Rightarrow {n^2} - n = 42 \\

\Rightarrow {n^2} - n - 42 = 0 \\

\Rightarrow {n^2} - 7n + 6n - 42 = 0 \\

\Rightarrow n(n - 7) + 6(n - 7) = 0 \\

\Rightarrow (n - 7)(n + 6) = 0 \\

$

$\therefore $ n=7 0r n=-6

We know that sides cannot be negative $\therefore n = 7$ is the required answer.

Hence, option B is the correct answer.

Note:-Whenever we face such a type of question the key concepts for solving the question is that you have to first choose the three sides from the n sides by selection method to form the triangle and then proceed according to the condition which is given in the question.

Complete step by step solution:

Given that,

${T_{n + 1}} - {T_n} = 21$

${T_{n + 1}}$ Can be written as ${}^{n + 1}{C_3}$

$ \Rightarrow {}^{(n + 1)}{C_3} - {}^n{C_3} = 21$

$ \Rightarrow {}^n{C_2} + {}^n{C_3} - {}^n{C_3} = 21$

$\because $ We know that ${}^{(n + 1)}{C_r} = {}^n{C_{r - 1}} + {}^n{C_r}$

$ \Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = 21$

$\because $We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$ \Rightarrow \dfrac{{n \times (n - 1) \times (n - 2)!}}{{2 \times (n - 2)!}} = 21$

$

\Rightarrow {n^2} - n = 42 \\

\Rightarrow {n^2} - n - 42 = 0 \\

\Rightarrow {n^2} - 7n + 6n - 42 = 0 \\

\Rightarrow n(n - 7) + 6(n - 7) = 0 \\

\Rightarrow (n - 7)(n + 6) = 0 \\

$

$\therefore $ n=7 0r n=-6

We know that sides cannot be negative $\therefore n = 7$ is the required answer.

Hence, option B is the correct answer.

Note:-Whenever we face such a type of question the key concepts for solving the question is that you have to first choose the three sides from the n sides by selection method to form the triangle and then proceed according to the condition which is given in the question.

Last updated date: 21st Sep 2023

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