# How many quadrilaterals formed from n-gon no adjacent sides

• member 428835
I think the easiest/surest way is to break it into cases.Let N211 be the number of ways of choosing four vertices such that two are adjacent and the other two isolated, etc.N1111+N211+N22+N31+N4=nC4N4=nN31=n(n-5)N22=n(n-5)/2N211 is the slightly tricky...
member 428835
Homework Statement
How many quadrilaterals can be formed from n-polygon (convex) given no adjacent sides of the polygon are used?
Relevant Equations
Nothing comes to mind.
My thought it to think of the orientation as a string: let V denote a selected vertex and E denote a vertex that's not selected. There are then 4 V's to distribute and ##n-4## E's to distribute. Then just choosing quadrilaterals without restriction we have ##n!/(4!(n-4)!)##. However, we want no adjacent sides. I'm thinking instead of selecting V,V,V,V we now position VE,VE,VE,VE with the remaining ##n-8## E's going anywhere without restriction. Thus the number of quadrilaters without adjacent sides is ##(n-8+4)!/((n-8)!4!)##. Is this correct, or am I missing something?

I have not investigated your answer but have you checked it for easy primitive cases, n=3,4,5,6 which would confirm/correct your answer ? For n=6 I find three quadrilaterals of such a condition.

member 428835
anuttarasammyak said:
I have not investigated your answer but have you checked it for easy primitive cases, n=3,4,5,6 which would confirm/correct your answer ? For n=6 I find three quadrilaterals of such a condition.
Don’t you mean n=8, 9, ...? (I get 2, 18, ...)

member 428835
haruspex said:
Don’t you mean n=8, 9, ...? (I get 2, 18, ...)
Did I take it wrong ?

anuttarasammyak said:
Did I take it wrong ?
View attachment 301590
Sorry, I read it as no two adjacent vertices. In my defence, I was influenced by the OP's attempt, which seems to have made the same mistake.

Fwiw, I now get 0, 3, 7, 38 for n=5, 6, 7, 8.

Last edited:
anuttarasammyak
haruspex said:
Sorry, I read it as no two adjacent vertices. In my defence, I was influenced by the OP's attempt, which seems to have made the same mistake.

Fwiw, I now get 0, 3, 7, 38 for n=5, 6, 7, 8.
You read it correct. So ##n > 8## to even get a solution. But you say there's a mistake; can you elaborate?

joshmccraney said:
You read it correct.
But it says "adjacent sides". For two adjacent sides to be part of the quadrilateral it would have to use three consecutive vertices, not just two. See the diagram in post #4.

haruspex said:
But it says "adjacent sides". For two adjacent sides to be part of the quadrilateral it would have to use three consecutive vertices, not just two. See the diagram in post #4.
Doesn't "no adjacent sides" imply you cannot have adjacent vertices? Diagram in 4 has these, and so is not the description I am looking for.

Example: vertices v1, v2, v3, v4, v5, v6, v7, v8, implies v1,v3,v5,v7 and v2,v4,v6,v8 are the only two solutions, since none of these quadrilateral sides are adjacent with the octogon.

joshmccraney said:
Doesn't "no adjacent sides" imply you cannot have adjacent vertices?
No. Look at e.g. the pink triangle in post #4. It uses two sides of the hexagon, but they are not adjacent in it.
Have you quoted the problem exactly as given to you? If it meant not use any sides of the polygon, why insert the word "adjacent"?

haruspex said:
No. Look at e.g. the pink triangle in post #4. It uses two sides of the hexagon, but they are not adjacent in it.
Have you quoted the problem exactly as given to you? If it meant not use any sides of the polygon, why insert the word "adjacent"?
Sorry, by adjacent I should have said "no same side" as the n-gon.

joshmccraney said:
Sorry, by adjacent I should have said "no same side" as the n-gon.
Ok.
I would try using inclusion-exclusion. Are you familiar with that?

member 428835
haruspex said:
Ok.
I would try using inclusion-exclusion. Are you familiar with that?
I am. But I thought my approach was correct?

joshmccraney said:
I am. But I thought my approach was correct?
With n=8 your formula gives 1. Shouldn’t it be 2? For n=9 your formula gives 5, but I count 9.

joshmccraney said:
Sorry, by adjacent I should have said "no same side" as the n-gon.
On polygon ABCD
a is number of vertexes of n-polygon between A and B
b is number of vertexes of n-polygon between B and C
c is number of vertexes of n-polygon between C and D
d is number of vertexes of n-polygon between D and A
I guess the number of cases satisfying
a+b+c+d=n-4
a,b,c,d ##\ge## 1
multiplied by n/4, which express A is on which vertex and remove count redundancy, would be the answer.In addition when the condition of "no adjacent sides of n-polygon" is also permitted, the cases of
$$a=0, db\neq0$$
$$b=0, ac\neq0$$
$$c=0, bd\neq0$$
$$d=0, ca\neq0$$
should be taken account also.

Last edited:
anuttarasammyak said:
On polygon ABCD
a is number of vertexes of n-polygon between A and B
b is number of vertexes of n-polygon between B and C
c is number of vertexes of n-polygon between C and D
d is number of vertexes of n-polygon between D and A
I guess the number of cases satisfying
a+b+c+d=n-4
a,b,c,d ##\ge## 1
multiplied by n/4, which express A is on which vertex and remove count redundancy, would be the answer.In addition when the condition of "no adjacent sides of n-polygon" is also permitted, the cases of
$$a=0, db\neq0$$
$$b=0, ac\neq0$$
$$c=0, bd\neq0$$
$$d=0, ca\neq0$$
should be taken account also.
A problem with that approach is symmetries. In your diagram in post #4, each of the rectangles would be counted twice. In a larger polygon there may be squares counted four times and quadrilaterals with no symmetries.

Prof B
joshmccraney said:
Homework Statement:: How many quadrilaterals can be formed from n-polygon (convex) given no adjacent sides of the polygon are used?
How is a quadrilateral "formed" from an n-gon? How is the convexity of the n-gon relevant? Why do you call it an n-polygon instead of an n-gon?

A possibly related question was already posed: https://www.physicsforums.com/threads/no-of-quadrilaterals-in-a-polygon.480827/

joshmccraney said:
Formed by selecting 4 vertices. I call it an n-gon in post 10.

I think the easiest/surest way is to break it into cases.
Let N211 be the number of ways of choosing four vertices such that two are adjacent and the other two isolated, etc.
N1111+N211+N22+N31+N4=nC4
N4=n
N31=n(n-5)
N22=n(n-5)/2
N211 is the slightly tricky one.

n ways to pick the 2. There are n-4 positions available to the two isolates. In n-5 ways these would also be adjacent, so N211=n(n-4C2-(n-5))

joshmccraney said:
Formed by selecting 4 vertices. I call it an n-gon in post 10.
There are 3 quadrilaterals with vertex set {A,B,C,D}. Do you want to count all 3 or only the convex one?

Prof B said:
There are 3 quadrilaterals with vertex set {A,B,C,D}. Do you want to count all 3 or only the convex one?
It doesn’t matter much because the ratio is 3:1 for all n.

haruspex said:
It doesn’t matter much because the ratio is 3:1 for all n.
Not necessarily. What if ABCD contains an edge of the n-gon but ACBD does not?

haruspex
anuttarasammyak said:
On polygon ABCD
a is number of vertexes of n-polygon between A and B
b is number of vertexes of n-polygon between B and C
c is number of vertexes of n-polygon between C and D
d is number of vertexes of n-polygon between D and A
I guess the number of cases satisfying
a+b+c+d=n-4
a,b,c,d ##\ge## 1
multiplied by n/4, which express A is on which vertex and remove count redundancy, would be the answer.
If the quadrilaterals are required to be convex then your analysis is correct. The number of cases is called a "composition" and you can find a formula here.

According to Georg Polya, understanding the question is the first step in problem solving.

Prof B said:
Not necessarily. What if ABCD contains an edge of the n-gon but ACBD does not?
Good point.

## 1. How do you determine the number of quadrilaterals formed from an n-gon with no adjacent sides?

The number of quadrilaterals formed from an n-gon with no adjacent sides can be determined by using the formula n*(n-3)/2, where n is the number of sides of the n-gon.

## 2. Can you provide an example to illustrate the formula for determining the number of quadrilaterals?

For example, if we have a pentagon (n=5), the number of quadrilaterals formed would be 5*(5-3)/2 = 5*2/2 = 5.

## 3. How does the number of quadrilaterals formed change as the number of sides of the n-gon increases?

As the number of sides of the n-gon increases, the number of quadrilaterals formed also increases. This is because there are more possible combinations of non-adjacent sides to form a quadrilateral.

## 4. Are there any exceptions to the formula for determining the number of quadrilaterals?

Yes, there are a few exceptions. If the n-gon has less than 4 sides, there will be no possible quadrilaterals. Additionally, if the n-gon has an odd number of sides, there will be one less quadrilateral than the formula suggests due to the inclusion of the center point.

## 5. Can this formula be applied to any polygon with non-adjacent sides?

Yes, this formula can be applied to any polygon with non-adjacent sides, as long as the polygon has at least 4 sides. The formula takes into account the number of sides of the polygon and the fact that the sides cannot be adjacent when forming a quadrilateral.

### Similar threads

• Precalculus Mathematics Homework Help
Replies
53
Views
6K
• Math Proof Training and Practice
Replies
25
Views
2K
• General Math
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Precalculus Mathematics Homework Help
Replies
2
Views
5K
• Special and General Relativity
Replies
9
Views
4K
• Math Proof Training and Practice
Replies
100
Views
8K
• Advanced Physics Homework Help
Replies
2
Views
2K
• Advanced Physics Homework Help
Replies
6
Views
2K
• Math Proof Training and Practice
Replies
46
Views
5K