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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 Verified
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.

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