In the last two posts, we discussed how to easily handle constraints in linear arrangements. Today we will discuss how to handle constraints in circular arrangements, which are actually even simpler to sort out. Let’s look at some examples.

Question 1: Seven people are to be seated at a round table. Andy and Bob don’t want to sit next to each other. How many seating arrangements are possible?

Solution: There are 7 people who need to be seated around a circular table. Number of arrangements in which 7 people can be seated around a circular table = (7-1)! = 6!

(If you are not sure how we got this, check out this post.)

We need to find the number of arrangements in which two of them do not sit together. Let us instead find the number of arrangements in which they will sit together. We will then subtract these arrangements from the total 6! arrangements. Consider Andy and Bob to be one unit. Now we need to arrange 6 units around a round table. We can do this in 5! ways. But Andy and Bob can swap places so we need to multiply 5! by 2.

Number of arrangements in which Andy and Bob do sit next to each other = 2*5!

So, number of arrangements in which Andy and Bob don’t sit next to each other = 6! – 2*5!

This is very similar to the way we handled such constraints in linear arrangements.

Question 2: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A cannot sit next to D and F at the same time. How many such arrangements are possible?

Solution: Total number of ways of arranging 6 people in a circle = 5! = 120

Now, A cannot sit next to D and F simultaneously.

Let’s first find the number of arrangements in which A sits between D and F. In how many of these 120 ways will A be between D and F? Let’s consider that D, A and F form a single unit. We make DAF sit on any three consecutive seats in 1 way and make other 3 people sit in 3! ways (since the rest of the 3 seats are distinct). But D and F can swap places so the number of arrangements will actually be 2*3! = 12

In all, we can make A sit next to D and F simultaneously in 12 ways.

The number of arrangements in which A is not next to D and F simultaneously is 120 – 12 = 108.

A slight variation of this question that would change the answer markedly is the following:

Question 3: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A can sit neither next to D nor next to F. How many such arrangements are possible?

Solution: In the previous question, A could sit next to D and F; the only problem was that A could not sit next to both of them at the same time. Here, A can sit next to neither D nor F. Generally, it is difficult to wrap your head around what someone cannot do. It is easier to consider what someone can do and go from there. A cannot sit next to D and F so he will sit next to two of B, C and E.

Let’s choose two out of B, C and E. In other words, let’s drop one of B, C and E. We can drop one of B, C and E in 3 ways (we can drop B or C or E). This means, we can choose two out of B, C and E in 3 ways (We will come back to choosing 2 people out of 3 when we work on combinations). Now, we can arrange the two selected people around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these three sit on any three consecutive seats in 1 way.

Number of ways of choosing two of B, C and E and arranging the chosen two with A = 3*2 = 6

The rest of the three people can sit in three distinct seats in 3! = 6 ways

Total number of ways in which A will sit next to only B, C or E (which means A will sit neither next to D nor next to F) = 6*6 = 36 ways

Now we will look at one last example.

Question 4: Six people are to be seated at a round table with seats arranged at equal distances. Andy and Bob don’t want to sit directly opposite to each other. How many seating arrangements are possible?

Solution: Directly opposite means that Andy and Bob cannot sit at the endpoints of the diameter of the circular table.

Total number of arrangements around the circular table will be (6-1)! = 5!

But some of these are not acceptable since Andy sits opposite Bob in these. Let us see in how many cases Andy doesn’t sit opposite Bob. Let’s say we make Andy sit first. He can sit at the table in 1 way since all the seats are exactly identical for him. Now there are 5 seats left but Bob can take a seat in only 4 ways since he cannot occupy the seat directly opposite Andy. Now there are 4 people left and 4 distinct seats left so they can be occupied in 4! ways.

Total number of ways of arranging the 6 people such that Andy does not sit directly opposite Bob = 1*4*4! = 96 arrangements.

Make sure you understand the logic used in this question. We will build up on it in the next post.

*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 GMAT prep for Veritas Prep and regularly participates in content development projects such as this blog!*

Amazing Post . I didnt learn arrangements so well in High School too

I guess most of us found arrangements really tricky in high school. Now I think they aren’t really that bad after all!

“Total number of ways of arranging the 6 people such that Andy does not sit next to Bob = 1*4*4! = 96 arrangements.”

I think you mean this to say ‘does not sit directly across from Bob.” Right?

Fantastic post btw, thank you very much!

Six people are to be seated at a round table with seats arranged at equal distances. Andy and Bob don’t want to sit directly opposite to each other. How many seating arrangements are possible?

Can I use this logic?

6 people can sit in 5! ways

Andy and B can not sit opposite to each other-Treat them as one person/set, since they will be on opposite ends of the diameter. It does not matter which one of the three diameters since arrangements are circular.

Rest can be seated in 4 seats. AB & rest4 can be seated in (5-1)! ways.

result=5!-4!=96

Yes it’s absolutely fine.

There are various other ways of looking at the solution too.

Method 2: Find the number of arrangements in which they are sitting together.

Make Andy sit on any seat in 1 way. Bob sits right opposite him in 1 way. Rest of the 4 people take 4 places in 4! ways. (pretty much what you have done)

Number of arrangements in which Any and Bob are not opposite each other = 5!-4! = 96

Method 3: Find the number of arrangements in which they don’t sit opposite each other

Make Andy sit on any seat in 1 way. Bob cannot sit opposite him so he has 4 options to choose from. Rest of the 4 people take 4 places in 4! ways.

Number of arrangements in which Andy and Bob are not opposite each other = 4*4! = 96

Hi Karishma,

I have used Method 3 of the above post. However, I have a small doubt.

We have fixed Andy and then we found out the number of ways B can sit such that B doesn’t sit opposite to A. This way we get 96 ways.

Don’t we need to fix B and then we check the number of ways A can sit such that A doesn’t sit opposite to B. This way also, we get 96 ways.

So total number of ways = 2*96= 192.

Please correct where I am going wrong.

Thanks

H

You are double counting. When we make Andy sit at any one place and then make Bob sit at the leftover acceptable 4 places, we count all the arrangements possible. If we make Bob sit first, the arrangements stay the same (it doesn’t matter who sits first. The end result is the same). So we are double counting.

Question 4: Six people are to be seated at a round table with seats arranged at equal distances. Andy and Bob don’t want to sit directly opposite to each other. How many seating arrangements are possible?

Total number of arrangements around the circular table will be (6-1)! = 5!

Let’s count the number of ways in which Andy and Bob can sit opposite to each other.We can make Andy sit first and make Bob sit right opposite to Andy.The number of ways to make this arrangement is one.Rest of the 4 people take 4 places in 4! ways.

Number of arrangements in which Andy and Bob are not opposite each other = 5! -1*4! = 96

Is the above logic correct?What if we make Bob sit first and make Andy sit right opposite,does the arrangement change?I’m not really sure whether the arrangement changes are not.I’m not able to visualize it and I get confused and come up with below answer.

Number of arrangements in which Andy and Bob are not opposite each other = 5! -2*4! = 72

Please help.

Yes. This method is correct. You work out the complement and subtract from the total.

The sequence in which you make the people sit is not important. What matters is the arrangement you see after everyone is seated. If Andy sits first, he takes any one of the 6 physical spots (which are identical) and Bob sits opposite him. Let’s number the spots for convenience. Say if Andy takes spot 1, Bob takes spot 4. If Andy takes spot 2, Bob takes spot 5 and so on… If Bob sits first, he takes any one of the 6 physical spots and Andy sits opposite him. If Bob takes spot 1, Andy takes spot 4. If Bob takes spot 4, Andy takes spot 1. But you have already accounted for all these positions when you made Andy sit first. Hence you do not multiply by 2. Answer remains 96 and not 72.

Total number of ways of arranging the 6 people such that Andy does not sit next to Bob = 1*4*4! = 96 arrangements.

Shouldn’t it be 3*4! as andy cannot sit in either position next to bob, so only three left. Method can be applied as well.

btw…..great post

Actually the question asks “directly opposite”. The perils of copy and paste! Updated.

Very insightful.

I have tried the example 4 in the following way.

Question 4: Six people are to be seated at a round table with seats arranged at equal distances. Andy and Bob don’t want to sit directly opposite to each other. How many seating arrangements are possible?

Solution:

Total number of arrangements around the circular table will be (6-1)! = 5!=120

As we know Andy will always sit opposite to one of the member.

Using symmetry, possible combination of Andy and any one of the remaining 5 members sitting opposite is 5!/5 = 120/5 = 24 arrangements.

So, total number of ways of arranging the 6 people such that Andy does not sit directly opposite Bob = 5! – (5!/5) = 120 -24= 96 arrangements.

Kindly comment.

Yes, your logic is correct. Out of the total arrangements after Andy occupies one place, Bob will take the place opposite him in (1/5)th of the cases since Bob is one of the 5 people for 5 places. In (4/5)th of the cases, Bob will not be opposite Andy i.e. in (4/5)*120 = 96

Hi karishma!

I am really a fan of your good work. I realise its an old post, nevermind giving a try….

for question 4, I understand what is done up there, problem i face is not being able to visualize 24 cases (120-96) where A and B sit opposite to each other.

its A1 B 4

A6 B 3

A5 B2

A4 B1

A3 B6

A2 B5

Please explain what am i missing.

Regards

Hi Karishma,

Thanks a lot for your post, it’s a great resource !! You make things very easy for people who don’t come from a quantitative background like me.

I have a problem understanding the difference in the number of ways we arrange the remaining people in an unacceptable arrangement.

In question 1 A and B cannot sit next to each other. I understand the logic here: we consider A & B as one unit so we have 6 remaining units to place. And because it’s a circular arrangement we can do so in 5! ways.

In question 2, we have 6 people and A cannot sit next to D and F at the same time. So we consider A,D and F one unit.I don’t understand why instead of having 5 remaining units for which there are 4! ways to sit, we have “3 people sit in 3! ways (since the rest of the 3 seats are distinct)'”.

I see the difference hinges on the fact that in the second problem the seats are distinct and not in the first but I don’t understand why that’s the case. What makes those seats distinct in one problem and not in the other ?

Thanks a lot for your help !

Lazure