Question
Question: Let \({T_n}\) denote the number of triangles which can be formed by using the vertices of a regular ...
Let Tn denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If Tn+1−Tn=21, then n equals
(A) 5 (B) 6 (C) 7 (D) None of theseSolution
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=r!×(p−r)!p!
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=r!×(p−r)!p!
Since, the triangle has three vertices, hence the value ofr will be 3. And polygon has n sides, therefore, p=n.
So, number of triangles formed can be written as
Tn=nC3=3!×(n−3)!n! ………………….. (1)
Now, a condition is given in the question,
Tn+1−Tn=21
We will put the values of Tn and Tn+1 in the above expression, we get
⇒Tn+1−Tn=21 ⇒n+1C3−nC3=21 .................(2)
Also, we know that n+1C3=nC3+nC2…………………..(3)
With this, putting the values in eq. (2), we get
⇒n+1C3−nC3=21 ⇒nC3+nC2−nC3=21 ⇒nC2=21 ⇒2!×(n−2)!n!=21 ⇒2×1×(n−2)!n.(n−1)(n−2)!=21 ⇒n(n−1)=42
Now, simplifying the above expression, we will get
⇒n2−n−42=0 ⇒n2−7n+6n−42=0 ⇒n(n−7)+6(n−7)=0 ⇒(n−7)(n+6)=0
This way, we will get two values ofnas
n=7,−6
Since the negative value ofnis 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.