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Question: Let \({{T}_{n}}\) be the number of all possible triangles formed by joining vertices of an \(n\) -si...

Let Tn{{T}_{n}} be the number of all possible triangles formed by joining vertices of an nn -sided regular polygon. If Tn+1−Tn=19{{T}_{n+1}}-{{T}_{n}}=19 then the value of nn is
A. 77
B. 55
C. 1010
D. 88

Explanation

Solution

In this problem we need to calculate the value of nn for the given conditions. We have given that the Tn{{T}_{n}} is the number of all possible triangles formed by joining the vertices of nn sided polygon. We know that to form a triangle we need to join 33 vertices. So we will calculate the value of Tn{{T}_{n}} by calculating the value of selecting 33 vertices from nn vertices. After having the value Tn{{T}_{n}} we can calculate the value of Tn+1{{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 Tn{{T}_{n}} is the number of all possible triangles formed by joining the vertices of nn sided polygon.
We know that a nnsided polygon has nn vertices and to form a triangle by joining the vertices of the polygon we need to join any 33 of the vertices. So the number of possible triangle that can be formed in the nn sided polygon is given by
Tn=nC3{{T}_{n}}={}^{n}{{C}_{3}}
From the above equation, the value of Tn+1{{T}_{n+1}} will be
Tn+1=n+1C3{{T}_{n+1}}={}^{n+1}{{C}_{3}}
Substituting the values of Tn{{T}_{n}} and Tn+1{{T}_{n+1}} in the given equation which is Tn+1−Tn=19{{T}_{n+1}}-{{T}_{n}}=19 , then we will get
n+1C3−nC3=10{}^{n+1}{{C}_{3}}-{}^{n}{{C}_{3}}=10
Using the formula nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} in the above equation, then we will have
(n+1)!(n+1−3)!3!−n!(n−3)!3!=10 ⇒(n+1)!(n−2)!3!−n!(n−3)!3!=10 \begin{aligned} & \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{aligned}
Writing the value (n+1)!\left( n+1 \right)! as (n+1)×(n+1−1)!=(n+1)n!\left( n+1 \right)\times \left( n+1-1 \right)!=\left( n+1 \right)n! and (n−2)!\left( n-2 \right)! as (n−2)×(n−2−1)!=(n−2)(n−3)!\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
(n+1)n!(n−2)(n−3)!3!−n!(n−3)!3!=10\dfrac{\left( n+1 \right)n!}{\left( n-2 \right)\left( n-3 \right)!3!}-\dfrac{n!}{\left( n-3 \right)!3!}=10
Taking n!(n−3)!3!\dfrac{n!}{\left( n-3 \right)!3!} as common from the two terms in LHS, and then we will have
n!(n−3)!3!(n+1n−2−1)=10 ⇒n!(n−3)!3!(n+1−n+2n−2)=10 ⇒n!(n−3)!3!(3n−2)=10 \begin{aligned} & \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{aligned}
Writing the value 3!3! as 3×(3−1)!=3×2!3\times \left( 3-1 \right)!=3\times 2! in the above equation, then we will get
n!(n−3)!3×2!(3n−2)=10 ⇒n!(n−2)(n−3)!2!=10 \begin{aligned} & \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{aligned}
We have the value (n−2)(n−3)!\left( n-2 \right)\left( n-3 \right)! as (n−2)!\left( n-2 \right)! and using the formula nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} in the above equation, then we will have
n!(n−2)!2!=10 ⇒nC2=10 \begin{aligned} & \dfrac{n!}{\left( n-2 \right)!2!}=10 \\\ & \Rightarrow {}^{n}{{C}_{2}}=10 \\\ \end{aligned}
We can write the value 1010 as 5C2{}^{5}{{C}_{2}} in the above equation, and then we will have
nC2=5C2{}^{n}{{C}_{2}}={}^{5}{{C}_{2}}
Equating on both sides of the above equation, then we will get
∴n=5\therefore n=5
Hence option – B is the correct answer.

Note: In this problem we have solved the equation n+1C3−nC3=10{}^{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+1Cr−nCr=nCr−1{}^{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.