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How many triangles can you form from 12 points if 5 of them lie on one line? Questions like this are a staple of permutations and combinations, and they reward one habit: work out what a figure needs, choose that many points, then subtract the choices that do not give a real figure. This post covers chords, diagonals, lines, triangles and rectangles, with numbers checked.
Where do these questions come from?
The JEE (Main) 2026 syllabus lists permutations and combinations, the meaning of P(n, r) and C(n, r), and "simple applications"; geometric counting is one of the commonest. NCERT Class 11 Chapter 6 introduces combinations with the question of how many chords can be drawn through 7 points on a circle, and asks the same for 21 points in an exercise. Check your exam year's syllabus. For the counting basics, see our guide to permutations and combinations.
What does each figure need?
A chord or line segment: 2 points, so ⁿC₂.
A triangle: 3 non-collinear points, so ⁿC₃ when no three are collinear.
A quadrilateral: 4 points, no three collinear, so ⁿC₄.
For 21 points on a circle, no three are collinear, so there are ²¹C₂ = 210 chords.
How many diagonals does a polygon have?
Joining any 2 of the n vertices gives ⁿC₂ segments. Of these, n are sides. So the number of diagonals is ⁿC₂ − n = n(n − 3)/2.
A decagon: 10 × 7/2 = 35 diagonals.
Reverse question: which polygon has 44 diagonals? Solve n(n − 3) = 88. n = 11 works, since 11 × 8 = 88.
Worked example: 12 points, 5 of them collinear
Lines. If no three were collinear, there would be ¹²C₂ = 66 lines. The 5 collinear points give ⁵C₂ = 10 pairs, but all 10 pairs give the same line. So subtract 10 and add 1 back: 66 − 10 + 1 = 57 lines.
Triangles. If no three were collinear, there would be ¹²C₃ = 220. Any 3 of the 5 collinear points give no triangle, and there are ⁵C₃ = 10 such choices. So 220 − 10 = 210 triangles.
We checked both numbers by placing 5 points on a line and 7 random points off it, and counting distinct lines and non-degenerate triangles by computer: 57 and 210.
Rectangles, squares and intersection points
Rectangles on a chessboard. An 8 × 8 board has 9 horizontal and 9 vertical lines. A rectangle needs 2 of each: ⁹C₂ × ⁹C₂ = 36 × 36 = 1296.
Squares on a chessboard. Count by size: 8² squares of side 1, 7² of side 2, and so on, giving 1 + 4 + ... + 64 = 204.
Intersection points of diagonals. In a convex n-gon where no three diagonals meet at one point, every 4 vertices give exactly one crossing of two diagonals. So there are ⁿC₄ crossings: 210 for a decagon.
Triangles with no side on the polygon. For a decagon there are ¹⁰C₃ = 120 triangles from its vertices, and the formula n(n − 4)(n − 5)/6 gives 50 of them sharing no side with the decagon. We confirmed 50 by listing all 120 by computer.
Why do students get these wrong?
Lockwood (2013) proposed a model of how students think about counting that links three things: the formula or expression they write, the counting process behind it, and the set of outcomes being counted. Geometric counting hides the set of outcomes inside a picture, which is why naming it out loud ("a triangle is a choice of 3 points, not all on the line") helps you pick the right formula. Eizenberg and Zaslavsky (2004) studied 14 undergraduates solving counting problems and describe verifying a solution as a major difficulty; of the 5 checking strategies they identified, 2 were more frequent and more helpful than the others.
Common mistakes and how to avoid them
Using ⁿC₃ for triangles when some points are collinear. Read the question twice for "collinear".
Forgetting to add 1 for the collinear line.
Counting sides as diagonals. Subtract n.
Counting only unit squares on a grid. Sum over all sizes.
Using ⁿC₄ for diagonal crossings when three diagonals meet at a point, which can happen in regular polygons. The question usually says "no three concurrent"; if not, be careful.
How can you check a count?
Use a small case you can draw. Four points with three collinear: the formula gives ⁴C₂ − ³C₂ + 1 = 4 lines and ⁴C₃ − ³C₃ = 3 triangles. Draw it and count. If the small case matches, the method is right; then trust the large numbers.
A practice routine
Do one question of each type this week: chords, diagonals, lines with collinear points, triangles with collinear points, rectangles on a grid.
Before each, write in words what one figure is: "a choice of 3 points not all on the line".
Check every method on a small case you can draw.
Mix these into sets with other counting questions so you must spot the selection yourself.
If you get stuck on why a formula subtracts or adds a term, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: count the triangles formed by 10 points of which 4 are collinear, then check your answer of 116 by the subtraction method.
