NDA Maths · Permutation & Combination

Geometric Counting

Counting lines, triangles, quadrilaterals, diagonals, and intersection points from a set of points or lines — combinations with a correction for collinear (degenerate) cases.

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Why this matters

Geometric counting is pure combination with one twist: collinear points make no triangle and concurrent/parallel lines lose intersections. Subtract the degenerate cases and these become routine.

Concept 1 of 1

Lines, triangles and polygons from points

Intuition

A line needs 2 points, a triangle 3, a quadrilateral 4 — so the raw counts are

^nC_2,\,^nC_3,\,^nC_4

. The correction: any set of collinear points that 'should' form a figure doesn't, so subtract those degenerate selections.

Definition

From $n$ points, no three collinear: lines $^nC_2$ , triangles $^nC_3$ , quadrilaterals $^nC_4$ . **If $k$ points are collinear:** subtract their degenerate selections — lines $^nC_2-^kC_2+1$ ; triangles $^nC_3-^kC_3$ . Diagonals of an $n$ -gon: $^nC_2-n=\dfrac{n(n-3)}{2}$ . Parallelograms from $m$ and $n$ parallel lines: $^mC_2\cdot{}^nC_2$ . Max intersection points of $n$ lines: $^nC_2$ .

Geometric counting from points and lines

\text{lines}=\,^nC_2 \qquad \text{triangles}=\,^nC_3 \qquad \text{diagonals}=\dfrac{n(n-3)}{2} \qquad \text{parallelograms}=\,^mC_2\cdot{}^nC_2

Worked example

How many triangles can be formed from 8 points, of which 3 are collinear?

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From the bank · past-year question

Example 1Permutation & CombinationMODERATE

For the following two (02) items: There are 8 points on a plane out of which 4 points are collinear.

How many triangles can be formed by joining these points?

[Q35 · Sep · 2025]

Collinear points form no triangle — subtract $^kC_3$

^nC_3

counts triangles only if no three points are collinear. If

k

of the points lie on one line, those

^kC_3

triples are degenerate (no triangle), so the answer is

^nC_3-\,^kC_3

. Likewise the diagonal formula

\dfrac{n(n-3)}{2}=\,^nC_2-n

subtracts the

n

sides from all

^nC_2

point-pairs.

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Summary — formulas & gotchas at a glance

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Formulas (1)

Lines, triangles and polygons from points
Geometric counting from points and lines
$\text{lines}=\,^nC_2 \qquad \text{triangles}=\,^nC_3 \qquad \text{diagonals}=\dfrac{n(n-3)}{2} \qquad \text{parallelograms}=\,^mC_2\cdot{}^nC_2$

Watch out for (1)

Collinear points form no triangle — subtract $^kC_3$
→ Lines, triangles and polygons from points

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