¹⁰C₃ = 120 triangles. No three points collinear since they're on a circle.

Find the Number of Triangles Formed by Joining 10 Points on a Circle

Question

Ten points lie on the circumference of a circle. How many triangles can be formed by joining these points?

Solution — Step by Step

A triangle needs exactly 3 non-collinear points. Since all 10 points lie on a circle, no three of them are collinear — any 3 points we pick will always form a valid triangle. This is the key insight that makes this problem clean.

We need to choose 3 points from 10. Order doesn’t matter here — picking points A, B, C gives the same triangle as picking C, A, B. So we use combinations, not permutations.

\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}

\frac{10 \times 9 \times 8}{6} = \frac{720}{6} = 120

The number of triangles = 120.

Why This Works

The core idea: every unique set of 3 points defines exactly one triangle. So counting triangles = counting 3-element subsets of our 10 points. That’s the definition of $\binom{10}{3}$ .

The circle condition is doing heavy lifting here. If some points were collinear (say, 3 points on the same straight line), those 3 points couldn’t form a triangle — we’d have to subtract those cases. On a circle, no three points are ever collinear, so every combination of 3 points is valid. Zero subtractions needed.

This is why circle-based PnC questions are usually cleaner than “n points in a plane, k of which are collinear” questions. The circle guarantees no collinearity for free.

Alternative Method

You can think of it as a sequential counting problem, then divide out the overcounting.

Choose the 1st vertex: 10 ways
Choose the 2nd vertex: 9 ways (any remaining point)
Choose the 3rd vertex: 8 ways

That gives $10 \times 9 \times 8 = 720$ ordered selections. But each triangle gets counted $3! = 6$ times (once for each ordering of its 3 vertices — ABC, ACB, BAC, BCA, CAB, CBA).

\text{Triangles} = \frac{720}{6} = 120

Same answer. This approach builds intuition for why we divide by $r!$ in the combination formula.

Quick mental check: $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$ . For $n = 10$ : $\frac{10 \cdot 9 \cdot 8}{6} = 120$ . Memorise this pattern — it shows up constantly in PnC questions on polygons and circles.

Common Mistake

Students often write $\binom{10}{3} \times 3!$ thinking they need to “arrange” the vertices. This gives 720 — the number of ordered triangles, not distinct triangles. Triangle ABC and triangle BCA are the same triangle. Use $\binom{10}{3}$ , not $P(10, 3)$ . Use permutations only when the problem asks for something order-dependent, like “how many ways to label the vertices”.

A secondary trap: students sometimes subtract “degenerate” cases without checking. On a circle, there are no degenerate triangles — every 3 points form a proper triangle. Don’t subtract anything. The answer is simply 120.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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