A polygon has n sides, the number of diagonals in this polygon is:

1. $\dfrac{1}{2}n(n - 1)$

2. $\dfrac{1}{2} \times n(n - 2)$

3. $\dfrac{1}{2} \times n(n - 3)$

4. $\dfrac{1}{2} \times n(n - 4)$

It is given that the polygon has n sides.

By joining any two opposite vertices of the polygon, we obtain diagonal of the polygon.

Total number of ways to join any two vertices (whether adjacent or not) is given by:

= ⁿC₂ - n

= $\dfrac{n(n - 1)}{2} - n$

= $\dfrac{n(n - 1)}{2} - \dfrac{n}{1}$

= $\dfrac{(n^2 - n)}{2} - \dfrac{2n}{2}$

= $\dfrac{(n^2 - n - 2n)}{2}$

= $\dfrac{(n^2 - 3n)}{2}$

= $\dfrac{n(n - 3)}{2}$

Hence, option 3 is the correct option.