Discussion

Explanation:

By selecting 4 vertices we can get exactly one pair of diagonals which will intersect each other. 

∴ The maximum number of points of intersection of the diagonals inside the polygon is same as selecting 4 vertices i.e., ⁿC₄.

⇒ ⁿC₄ = 330

⇒ $\frac{n(n-1)(n-2)(n-3)}{4\times 3\times 2\times 1}$ = 330

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 3 × 4 × 3 × 2

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 9 × 8

⇒ n = 11

Hence, 11.

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