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 ${T_{n + 1}} - {T_n} = 21$, then $n$ equals
\[
  {\text{(A) }}5 \\
  {\text{(B) 6}} \\
  {\text{(C) 7}} \\
  {\text{(D) None of these}} \\
 \]

Answer 1

Hint:This problem is related to the Permutation and Combination. It can be observed that to make a triangle within the polygon, the verticals can be reused. Hence, combination formula will be used. The standard formula is ${}^p{C_r} = \dfrac{{p!}}{{r! \times (p - r)!}}$

Complete step-by-step answer:
In the problem, it is to find the number of triangles that can be made within the polygon using and re-using the verticals. To calculate, we will use combination formula as
${}^p{C_r} = \dfrac{{p!}}{{r! \times (p - r)!}}$
Since, the triangle has three vertices, hence the value of$r$ will be 3. And polygon has $n$ sides, therefore, $p = n$.
So, number of triangles formed can be written as
${T_n} = {}^n{C_3} = \dfrac{{n!}}{{3! \times (n - 3)!}}$ ………………….. (1)
Now, a condition is given in the question,
${T_{n + 1}} - {T_n} = 21$
We will put the values of ${T_n}$ and ${T_{n + 1}}$ in the above expression, we get
$
   \Rightarrow {T_{n + 1}} - {T_n} = 21 \\
   \Rightarrow {}^{n + 1}{C_3} - {}^n{C_3} = 21{\text{ }}.................{\text{(2)}} \\
 $
Also, we know that ${}^{n + 1}{C_3} = {}^n{C_3} + {}^n{C_2}………………….. (3)$
With this, putting the values in eq. (2), we get
$
   \Rightarrow {}^{n + 1}{C_3} - {}^n{C_3} = 21 \\
   \Rightarrow {}^n{C_3} + {}^n{C_2} - {}^n{C_3} = 21 \\
   \Rightarrow {}^n{C_2} = 21 \\
   \Rightarrow \dfrac{{n!}}{{2! \times (n - 2)!}} = 21 \\
   \Rightarrow \dfrac{{n.(n - 1)(n - 2)!}}{{2 \times 1 \times (n - 2)!}} = 21 \\
   \Rightarrow n(n - 1) = 42 \\
 $
Now, simplifying the above expression, we will get
$
   \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 \\
 $
This way, we will get two values of$n$as
$n = 7, - 6$
Since the negative value of$n$is not acceptable because no. of verticals of polygon cannot be less than zero. Hence,
 $n = 7$

So, the correct answer is “Option C”.

Note:It is noted here that we have used eq. (3) to simplify eq. (2). If we would have not used that relationship, we would be ending with an equation of order 3 and solving a third order equation would take too much time. So it’s a good practice to simplify the equations to lower order than higher one. Higher order equations are, manually, tough to solve and also takes time.