1. **Problem statement:** Consider a circle with $n$ points marked on it, where $n \geq 2$. We want to find how many different chords can be drawn by connecting two of these $n$ points. 2. **Formula used:** The number of chords formed by connecting two points out of $n$ points is the number of combinations of $n$ points taken 2 at a time, which is given by: $$\binom{n}{2} = \frac{n(n-1)}{2}$$ 3. **Explanation:** - Each chord is uniquely determined by choosing 2 distinct points on the circle. - Since order does not matter (chord between point A and B is the same as between B and A), we use combinations, not permutations. 4. **Intermediate work:** - Calculate the number of ways to choose 2 points from $n$ points: $$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}$$ 5. **Final answer:** The number of different chords that can be drawn is: $$\boxed{\frac{n(n-1)}{2}}$$