Combinations

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The number of diagonals in a polygon with n sides is

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n(n-1)/2

n(n-3)/2

n(n-2)/2

n(n+1)/2

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About This Question

Subject: mathematics
Chapter: permutations and combinations
Topic: combinations
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

n(n-3)/2

Total Segments: A polygon with $n$ vertices allows $(2 n)$ total ways to join any two vertices.
- Total Segments $= \frac{n ( n - 1 )}{2}$ .
Filter Segments: These segments consist of either sides or diagonals.
Subtract Sides: There are $n$ sides in an $n$ -sided polygon.
- Diagonals $= \frac{n ( n - 1 )}{2} - n$ .
Simplification:
- $\frac{n ^{2} - n - 2 n}{2} = \frac{n ^{2} - 3 n}{2} = \frac{n ( n - 3 )}{2}$ .
Example: A square ( $n = 4$ ) has $\frac{4 ( 4 - 3 )}{2} = 2$ diagonals.

This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of combinations. It appeared in the 2025 exam.

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