Question

Let $t_{n}$ be the number of possible quadrilaterals formed by joining vertices of an n sided regular polygon. If $t_{n + 1} - t_{n} = 56$, then the number of triangles formed by joining vertices of the polygon is

• A. $120$
• B. $84$
• C. $56$
• D. $35$

Answer 1

Answer: (C)

As $t_{n}$ is the number of ways of selecting four vertices from n sided regular polygon
$\therefore t_{n} = \,{}^{n} C_{4} \therefore t_{n + 1} - t_{n} = 56$
$\Rightarrow \,{}^{n +1}C_{4} - \,{}^{n}C_{4} = 56$
$\Rightarrow \,{}^{n}C_{4} + \,{}^{n}C_{3} - \,{}^{n}C_{4} = 56$
$\Rightarrow \,{}^{n}C_{3} = 56$
$\Rightarrow \frac{n \left(n-1\right) \left(n-2\right)}{3\cdot2\cdot1} = 7 \times 8$
$\Rightarrow n \left(n -1\right) \left(n -2\right) = 8 \cdot7\cdot6$
$\Rightarrow n = 8$
$\therefore $ Required number of triangles
$= \,{}^{n}C_{3} = \,{}^{8}C_{3} = \frac{8 \cdot7\cdot6}{3 \cdot2 \cdot1} = 56$