Twelve distinct points lie on a circle.\nHow many different cyclic quadrilaterals can be formed by joining these points?

Introduction / Context:
This problem tests knowledge of combinations and the idea that any four distinct points on a circle determine a unique cyclic quadrilateral. The key realisation is that we are selecting sets of four points regardless of their order, so combinations are used instead of permutations.

Given Data / Assumptions:

• There are 12 distinct points on the circumference of a circle.
• Every set of 4 distinct points on the circle determines exactly one cyclic quadrilateral.
• No three points are collinear except at the circle, so we do not have degeneracies.
• Order of vertices in naming the quadrilateral does not matter.

Concept / Approach:
To form a quadrilateral, we need to choose 4 distinct vertices. Because the quadrilateral is determined entirely by which 4 points are selected, we just count how many 4 element subsets can be chosen from the 12 points. This is exactly the combination C(12,4). No extra factor is needed for different orderings because the shape is the same regardless of the direction in which vertices are listed.

Step-by-Step Solution:
Number of ways to choose 4 vertices from 12 points = C(12,4). Compute C(12,4) = 12! / (4! * 8!). Simplify using product form: C(12,4) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1). First compute the numerator: 12 * 11 = 132, 10 * 9 = 90, then 132 * 90 = 11880. Compute the denominator: 4 * 3 * 2 * 1 = 24. Now divide 11880 / 24. 24 * 495 = 11880, so C(12,4) = 495.

Verification / Alternative check:
An approximate way to confirm: C(12,4) should be less than 12^4 (which is 20736) and more than 12^3 (which is 1728) divided by a small factor, so a value in the low hundreds is expected. Also, C(12,4) equals C(12,8) by symmetry, and both correspond to choosing a relatively small subset from 12. The exact arithmetic 11880 / 24 cleanly gives 495, confirming the calculation.

Why Other Options Are Wrong:

• 500: A rounded figure, not a true combination result.
• 490: Close but not equal to C(12,4), reflects a small arithmetic error.
• 540: Too large; there is no combination of integer factorials from this problem that yields 540.

Common Pitfalls:
Students sometimes think that the circular nature of the points changes the count for quadrilaterals, but for selecting vertices it does not, because any choice of 4 points on the circle gives a unique cyclic quadrilateral. A more serious mistake is to use permutations P(12,4) instead of combinations, which would overcount arrangements of the same quadrilateral many times. The rule is simple: when only which points are chosen matters, not their order, always use combinations.

Final Answer:
The number of distinct cyclic quadrilaterals that can be formed is 495.