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Question: Let \({T_n}\) denote the number of triangles which can be formed by using the vertices of a regular ...

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

(A) 5 (B) 6 (C) 7 (D) None of these  {\text{(A) }}5 \\\ {\text{(B) 6}} \\\ {\text{(C) 7}} \\\ {\text{(D) None of these}} \\\
Explanation

Solution

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 pCr=p!r!×(pr)!{}^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
pCr=p!r!×(pr)!{}^p{C_r} = \dfrac{{p!}}{{r! \times (p - r)!}}
Since, the triangle has three vertices, hence the value ofrr will be 3. And polygon has nn sides, therefore, p=np = n.
So, number of triangles formed can be written as
Tn=nC3=n!3!×(n3)!{T_n} = {}^n{C_3} = \dfrac{{n!}}{{3! \times (n - 3)!}} ………………….. (1)
Now, a condition is given in the question,
Tn+1Tn=21{T_{n + 1}} - {T_n} = 21
We will put the values of Tn{T_n} and Tn+1{T_{n + 1}} in the above expression, we get
Tn+1Tn=21 n+1C3nC3=21 .................(2)  \Rightarrow {T_{n + 1}} - {T_n} = 21 \\\ \Rightarrow {}^{n + 1}{C_3} - {}^n{C_3} = 21{\text{ }}.................{\text{(2)}} \\\
Also, we know that n+1C3=nC3+nC2..(3){}^{n + 1}{C_3} = {}^n{C_3} + {}^n{C_2}………………….. (3)
With this, putting the values in eq. (2), we get
n+1C3nC3=21 nC3+nC2nC3=21 nC2=21 n!2!×(n2)!=21 n.(n1)(n2)!2×1×(n2)!=21 n(n1)=42  \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
n2n42=0 n27n+6n42=0 n(n7)+6(n7)=0 (n7)(n+6)=0  \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 ofnnas
n=7,6n = 7, - 6
Since the negative value ofnnis not acceptable because no. of verticals of polygon cannot be less than zero. Hence,
n=7n = 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.