Solveeit Logo
Login

Question

Question: Let \({T_n}\) be the number of all possible triangles formed by joining vertices of an n-sided regul...

Let Tn{T_n} be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10, then the value of n is?
(A). 5
(B). 10
(C). 8
(D). 7

Explanation

Solution

Start by forming the term by selecting 3 sides for triangle out of n sides of a polygon, follow the same for (n+1) sided polygon. Substitute the value in the given equation and apply the relevant formulas of combination in order to get the desired value of n (neglect any negative value).

Complete step-by-step answer :
Step by step solution
Given, Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10
Tn{T_n}is the number of all possible triangles formed by joining vertices of n sided polygon.
And we know that for any triangle to be formed we need 3 sides of any polygon. Therefore , selecting 3 sides out of n sides can be done in nC3{}^n{C_3}ways.
∴Tn=nC3\therefore {T_n} = {}^n{C_3}
Similarly , if we have (n+1) sided polygon ,we can select 3 sides out of it in n+1C3{}^{n + 1}{C_3} ways.
∴Tn+1=n+1C3\therefore {T_{n + 1}} = {}^{n + 1}{C_3}
Now , let us solve for the relation Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10and find out n value.
Tn+1−Tn=10 ⇒n+1C3−nC3=10  {T_{n + 1}} - {T_n} = 10 \\\ \Rightarrow {}^{n + 1}{C_3} - {}^n{C_3} = 10 \\\
And we know , nCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r} . Applying this formula in the equation , we get
⇒n+1C3−nC3=10 ⇒nC2+nC3−nC3=10 ⇒nC2=10  \Rightarrow {}^{n + 1}{C_3} - {}^n{C_3} = 10 \\\ \Rightarrow {}^n{C_2} + {}^n{C_3} - {}^n{C_3} = 10 \\\ \Rightarrow {}^n{C_2} = 10 \\\
We know that nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, Applying this formula
⇒n!2!(n−2)!=10 ⇒n×(n−1)×(n−2)!2!(n−2)!=10 ⇒n×(n−1)=20 ⇒n2−n−20=0  \Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = 10 \\\ \Rightarrow \dfrac{{n \times (n - 1) \times (n - 2)!}}{{2!\left( {n - 2} \right)!}} = 10 \\\ \Rightarrow n \times (n - 1) = 20 \\\ \Rightarrow {n^2} - n - 20 = 0 \\\
Splitting the middle term ,we get
n2−5n+4n−20=0 ⇒n(n−5)+4(n−5)=0 ⇒(n−5)(n+4)=0 ⇒n=5,−4  {n^2} - 5n + 4n - 20 = 0 \\\ \Rightarrow n(n - 5) + 4(n - 5) = 0 \\\ \Rightarrow (n - 5)(n + 4) = 0 \\\ \Rightarrow n = 5, - 4 \\\
As n can never have a negative value , so we’ll neglect -4.
Therefore , the value of n=5.
So , option A is the correct answer.

Note : All the formulas used in combination must be well known , as such formulas are very important and help in solving the question faster. Also, attention needs to be given while substituting the values and solving the quadratic equations if any. Any negative value of n or r must be neglected as they can never be negative.