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JEE Mains & Advanced Reading Time: 12 min 3 Core Concepts

Geometrical Combinatorics: Lines, Triangles, and Polygons

Master the core concepts of selection minus degenerate cases with formulas, examples, and JEE-level practice problems.

3
Key Formulas
100%
JEE Relevance
8+
Examples
15min
Avg. Solve Time

Why Geometrical Combinatorics Matters for JEE

Geometrical combinatorics bridges combinatorial mathematics with geometry, appearing frequently in both JEE Mains and Advanced. These problems test your understanding of:

  • Selection principles applied to geometric configurations
  • Elimination of degenerate cases (collinear points, etc.)
  • Systematic counting approaches for complex geometric figures
  • Formula application with careful condition checking
Concept 1 Medium

Number of Straight Lines from n Points

Finding maximum and actual number of straight lines formed by n points in a plane.

Key Formula:

Number of lines = $^nC_2 - ^mC_2 + 1$

Where m points are collinear (lie on the same straight line)

Understanding the Formula:

Maximum lines from n points = $^nC_2$ (if no three points are collinear)

When m points are collinear:

  • We lose $^mC_2 - 1$ lines (all combinations of m points give only 1 line instead of mC₂ lines)
  • Add 1 for the single line through all collinear points

Example 1: 10 points, no three collinear

Step 1: Maximum lines = $^{10}C_2 = \frac{10 \times 9}{2} = 45$

Step 2: Since no three collinear, all 45 lines are distinct

Answer: 45 straight lines

Example 2: 8 points with 5 collinear

Step 1: Maximum lines if no collinear = $^8C_2 = 28$

Step 2: Apply formula: Lines = $^8C_2 - ^5C_2 + 1$

Step 3: Calculate: $28 - 10 + 1 = 19$

Verification: From 5 collinear points, we get only 1 line instead of 10 possible lines

Answer: 19 straight lines

Concept 2 Hard

Number of Triangles from n Points

Finding triangles formed by n points with various collinear conditions.

Key Formula:

Number of triangles = $^nC_3 - ^mC_3$

Where m points are collinear (cannot form triangles)

Understanding the Formula:

Maximum triangles from n points = $^nC_3$ (if no three points are collinear)

When m points are collinear:

  • We subtract $^mC_3$ because any 3 points from m collinear points cannot form a triangle
  • The formula eliminates these "degenerate" cases

Example 1: 12 points, no three collinear

Step 1: Maximum triangles = $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

Answer: 220 triangles

Example 2: 10 points with 4 collinear

Step 1: Maximum triangles = $^{10}C_3 = 120$

Step 2: Apply formula: Triangles = $^{10}C_3 - ^4C_3$

Step 3: Calculate: $120 - 4 = 116$

Verification: From 4 collinear points, any 3 points give 0 triangles (4C₃ = 4 invalid combinations)

Answer: 116 triangles

Example 3: 15 points with two sets: 5 collinear and 4 collinear

Step 1: Maximum triangles = $^{15}C_3 = 455$

Step 2: Subtract invalid triangles from both collinear sets

Step 3: Triangles = $^{15}C_3 - ^5C_3 - ^4C_3$

Step 4: Calculate: $455 - 10 - 4 = 441$

Answer: 441 triangles

Concept 3 Medium

Number of Diagonals in an n-sided Polygon

Finding diagonals in convex polygons and special cases.

Key Formula:

Number of diagonals = $^nC_2 - n = \frac{n(n-3)}{2}$

Total lines minus sides of the polygon

Understanding the Formula:

From n vertices, total lines = $^nC_2$

We subtract n because:

  • n of these lines are sides of the polygon
  • The remaining are diagonals

Example 1: Hexagon (6-sided polygon)

Step 1: Using formula: Diagonals = $\frac{6(6-3)}{2} = \frac{6 \times 3}{2} = 9$

Step 2: Verification: Total lines = $^6C_2 = 15$, minus 6 sides = 9 diagonals

Answer: 9 diagonals

Example 2: Decagon (10-sided polygon)

Step 1: Using formula: Diagonals = $\frac{10(10-3)}{2} = \frac{10 \times 7}{2} = 35$

Answer: 35 diagonals

Example 3: Number of diagonals that can be drawn from one vertex

Step 1: From one vertex, we can draw diagonals to all other vertices except:

• The vertex itself

• The two adjacent vertices (these would be sides)

Step 2: Diagonals from one vertex = $n - 3$

Example: For octagon (n=8): $8 - 3 = 5$ diagonals from one vertex

🚀 Problem-Solving Strategies

For Lines & Triangles:

  • Always check for collinear points first
  • Remember: 2 points always give 1 line
  • 3 collinear points give 0 triangles
  • For multiple collinear sets, subtract each separately

For Polygons:

  • Diagonals = Total lines − Sides
  • From one vertex: n − 3 diagonals
  • Regular polygons have equal diagonal lengths
  • Convex polygons have all diagonals inside

Advanced Applications Available

Includes intersection points of diagonals, triangles in polygons, and special geometrical configurations

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. Find number of lines from 12 points with exactly 4 collinear

2. How many triangles from 15 points with 6 collinear?

3. Find diagonals in a 12-sided polygon and from one vertex

4. 20 points with two collinear sets: 5 and 4 points. Find triangles.

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