In today’s post, we will give you a question with two solutions and two different answers. You have to find out the correct answer and explain why the other is wrong. But before we do that, let’s give you some background.

Given an n sided polygon, how many diagonals will it have?

An n sided polygon has n vertices. If you join every distinct pair of vertices you will get nC2 lines. These nC2 lines account for the n sides of the polygon as well as for the diagonals.

So the number of diagonals is given by nC2 – n.

nC2 – n = n(n-1)/2 – n = **n(n – 3)/2**

Taking quick examples:

**Example 1**: How many diagonals does a polygon with 25 sides have?

No. of diagonals = n(n – 3)/2 = 25*(25 – 3)/2 = 275

**Example 2**: How many diagonals does a polygon with 20 sides have, if one of its vertices does not send any diagonal?

The number of diagonals of a 20 sided figure = 20*(20 – 3)/2 = 170

But one vertex does not send any diagonals. Each vertex makes a diagonal with (n-3) other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right. With all other vertices, it makes a diagonal. So we need to remove 20 – 3 = 17 diagonals from the total.

Total number of diagonals if one vertex does not make any diagonals = 170 – 17 = 153 diagonals.

We hope everything done till now makes sense. Now let’s go on to the part which seems to make no sense at all!

**Question**: How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

**Answer**: We will use two different methods to solve this question:

Method 1: Using the formula discussed above

Number of diagonals in a polygon of 18 sides = 18*(18 – 3)/2 = 135 diagonals

Each vertex makes a diagonal with n-3 other vertices – as discussed before.

So each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any diagonals = 135 – 15*3 = 90 diagonals.

Method 2:

The polygon has a total of 18 vertices. 3 vertices do not participate so we need to make all diagonals that we can with 15 vertices.

Number of lines you can make with 15 vertices = 15C2 = 15*14/2 = 105

But this 105 includes the sides as well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices do not participate, 4 sides will not be formed. 15 vertices will have 14 sides which will be a part of the 105 we calculated before.

Total number of diagonals if 3 vertices do not send any diagonals = 105 – 14 = 91

Note that the two answers do not match. Method 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct. Your job is to tell us which method is correct and why the other method is incorrect.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

I was wrong, please delete my last post.

Method 1 is wrong, because of 3 adjacent vertices, which means that one diagonal will be made by two of them. Therefore, when you subtract diagonals using (18-3)*3, you will subtract the diagonal 2 times.

Karishma,

The problem with Method 1 with three vertices not sending a diagonal is that you end up counting for one case an additional diagonal. Let me explain. Consider a pentagon, ABCDE discounting diagonals from A,B and C. So, valid outgoing diagonals minus adjacent sides would be AC and AD for A, BD and BE for B,CA and CE for C – there’s a repeat in AC and CA,there-by generating the incorrect 90 count.

Thanks,

Sriram.

Perfect!