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:
1. 5
2. 7
3. 6
4. 4

Answer 1

Correct Answer - Option 2 : 7

Concept:

• The number of ways in which r distinct objects can be selected from a group of n distinct objects is nCr = \(\rm \dfrac {n!}{r!(n-r)!}\).
• n! = n × (n -1) × ... × 3 × 2 × 1 = n(n - 1)!.
• 0! = 1.

 

Calculation:

Since, all the points (vertices) of a regular polygon will be definitely non-collinear, a triangle can always be formed by selecting any three vertices.

∴ The number of triangles which can be formed from n non-collinear points = Number of ways in which three distinct points can be selected out of n distinct points.

⇒ T_n = ⁿC₃ = \(\rm \dfrac {n!}{3!(n-3)!}=\dfrac {n!}{6(n-3)!}\).

And, Tn+1 = n+1C3 = \(\rm \dfrac {(n+1)!}{3![(n+1)-3]!}=\dfrac {(n+1)!}{6(n-2)!}\).

Now, Tn+1 - Tn = 21.

⇒ \(\rm \dfrac {(n+1)!}{6(n-2)!}-\dfrac {n!}{6(n-3)!}=21\)

⇒ \(\rm \dfrac {(n+1)n(n-1)(n-2)!}{6(n-2)!}-\dfrac {n(n-1)(n-2)(n-3)!}{6(n-3)!}=21\)

⇒ (n + 1)n(n - 1) - n(n - 1)(n - 2) = 126

⇒ n(n - 1)[(n + 1 - (n - 2)] = 126

⇒ n(n - 1)(3) = 126

⇒ n² - n - 42 = 0

⇒ n2 - 7n + 6n - 42 = 0

⇒ n(n - 7) + 6(n - 7) = 0

⇒ (n - 7)(n + 6) = 0

⇒ n - 7 = 0 OR n + 6 = 0

⇒ n = 7 OR n = -6.

Since n is the number of sides of a polygon, it cannot be negative.

∴ The value of n is 7.