How many possible arrangements can be made if 5 people be arranged to sit around a circular table and 2 people must sit together?

In this section you will learn to

Count the number of possible permutations of items arranged in a circle
Count the number of possible permutations when there are repeated items

In this section we will address the following two problems.

In how many different ways can five people be seated in a circle?
In how many different ways can the letters of the word MISSISSIPPI be arranged?

The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements.

Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! or 6 ways. Our next problem is to see how many ways these people can be seated in a circle. We draw a diagram.

It happens that there are only two ways we can seat three people in a circle, relative to each other’s positions. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. We are interested in the position of each person in relation to the others. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation. So in circular permutations, the first person is considered a place holder, and where he sits does not matter.

The number of permutations of \(n\) elements in a circle is \((n-1)!\)

In how many different ways can five people be seated at a circular table?

Solution

We have already determined that the first person is just a place holder. Therefore, there is only one choice for the first spot. We have

So the answer is 24.

In how many ways can four couples be seated at a round table if the men and women want to sit alternately?

Solution

We again emphasize that the first person can sit anywhere without affecting the permutation.

So there is only one choice for the first spot. Suppose a man sat down first. The chair next to it must belong to a woman, and there are 4 choices. The next chair belongs to a man, so there are three choices and so on. We list the choices below.

So the answer is 144.

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labeling the letters as follows.

\[E_1LE_2ME_3NT \nonumber \]

Since all the letters are now different, there are 7! different permutations.

Let us now look at one such permutation, say

\[LE_1ME_2NE_3T \nonumber \]

Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3! or 6 such arrangements. We list them below.

\begin{aligned} &\mathrm{LE}_{1} \mathrm{ME}_{2} \mathrm{NE}_{3} \\ &\mathrm{LE}_{1} \mathrm{ME}_{3} \mathrm{NE}_{2} \\ &\mathrm{LE}_{2} \mathrm{ME}_{1} \mathrm{NE}_{3} \mathrm{T} \\ &\mathrm{LE}_{2} \mathrm{ME}_{3} \mathrm{NE}_{1} \mathrm{T} \\ &\mathrm{LE}_{3} \mathrm{ME}_{2} \mathrm{NE}_{1} \mathrm{T} \\

& \mathrm{LE}_{3} \mathrm{ME}_{I} \mathrm{NE}_{2} \mathrm{T} \end{aligned}

Because the E's are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.

Let us suppose there are n different permutations of the letters ELEMENT.

Then there are \(n \cdot 3!\) permutations of the letters \(E_1LE_2ME_3NT\).

But we know there are 7! permutations of the letters \(E_1LE_2ME_3NT\).

Therefore, \(n \cdot 3! = 7!\)

Or \(n = \frac{7!}{3!}\).

This gives us the method we are looking for.

The number of permutations of n elements taken \(n\) at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, is

\[\frac{n !}{r_{1} ! r_{2} ! \ldots r_{k} !} \nonumber \]

Find the number of different permutations of the letters of the word MISSISSIPPI.

Solution

The word MISSISSIPPI has 11 letters. If the letters were all different there would have been 11! different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike.

So the answer is \(\frac{11!}{4!4!2!} = 34,650\).

If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there?

Solution

Again, we have permutations with similar elements.

We are looking for permutations for the letters HHHHTT.

The answer is \(\frac{6!}{4!2!} = 15\).

In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row?

Solution

Assuming that all nickels are similar, all dimes are similar, and all quarters are similar, we have permutations with similar elements. Therefore, the answer is

\[\frac{9 !}{4 ! 3 ! 2 !}=1260 \nonumber \]

A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done?

Solution

This means that each sales person gets 5 clients. The problem can be thought of as an ordered partitions problem. In that case, using the formula we get

\[\frac{20 !}{5 ! 5 ! 5 ! 5 !}=11,732,745,024 \nonumber \]

A shopping mall has a straight row of 5 flagpoles at its main entrance plaza. It has 3 identical green flags and 2 identical yellow flags. How many distinct arrangements of flags on the flagpoles are possible?

Solution

The problem can be thought of as distinct permutations of the letters GGGYY; that is arrangements of 5 letters, where 3 letters are similar, and the remaining 2 letters are similar:

\[ \frac{5 !}{3 ! 2 !} = 10 \nonumber \]

Just to provide a little more insight into the solution, we list all 10 distinct permutations:

GGGYY, GGYGY, GGYYG, GYGGY, GYGYG, GYYGG, YGGGY, YGGYG, YGYGY, YYGGG

We summarize.

1. Circular Permutations

The number of permutations of n elements in a circle is

\[(n -1)! \nonumber \]

2. Permutations with Similar Elements

The number of permutations of n elements taken n at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, such that \(\mathrm{n}=\mathrm{r}_{1}+\mathrm{r}_{2}+\ldots+\mathrm{r}_{\mathrm{k}}\) is

\[\frac{n !}{r_{1} ! r_{2} ! \dots r_{k} !} \nonumber \]

This is also referred to as ordered partitions.

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At a dinner party, 5 people are to be seated around a circular table. [#permalink]

Updated on: 01 Nov 2018, 01:59

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E. 120

Originally posted by marcodonzelli on 26 Dec 2007, 07:16.
Last edited by Bunuel on 01 Nov 2018, 01:59, edited 2 times in total.

Renamed the topic and edited the question.

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At a dinner party, 5 people are to be seated around a circular table. [#permalink]

24 Mar 2013, 01:21

Val1986 wrote:

At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?A. 5B. 10C. 24D. 32

E. 120

We have a case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\).The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:\(R = \frac{n!}{n} = (n-1)!\)"\((n-1)!=(5-1)!=24\)Answer: C.Questions about this concept to practice:

http://gmatclub.com/forum/seven-men-and ... 92402.html

http://gmatclub.com/forum/a-group-of-fo ... 88604.html
http://gmatclub.com/forum/the-number-of ... 94915.html
http://gmatclub.com/forum/in-how-many-d ... 02187.html
http://gmatclub.com/forum/4-couples-are ... 31048.html
http://gmatclub.com/forum/at-a-party-5- ... 04101.html
http://gmatclub.com/forum/seven-family- ... 02184.html
http://gmatclub.com/forum/seven-men-and ... 11473.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/seven-men-and ... 98185.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/find-the-numb ... 06919.htmlHope it helps. _________________

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

07 Jan 2008, 01:24

sort cut for any circular seating arrangements:
(n-1)! =(5-1)! = 4! = 4*3*2*1 = 24

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

26 Dec 2007, 07:33

C let 1,2,3,4,5 are people. 1. we fix position of 1 2. we have 4*3=12 possible positions for left and right neighbors of 1. 3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.

Therefore, N=12*2=24

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

26 Dec 2007, 09:49

OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

26 Dec 2007, 10:08

marcodonzelli wrote:

OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!

let 1,2,3,4,5 are people.

"_ _ _ _ _" - positions

1. we fix position of 1

"_ _ 1 _ _"

2. we have 4*3=12 possible positions for left and right neighbors of 1.

"_ x 1 _ _" x e {2,3,4,5}. 4 variants
"_ x 1 y _" y e {(2,3,4,5} - {x}. 3 variants

total number of variants is 4*3=12 3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.

or"a x 1 y _" a e {(2,3,4,5} - {x,y}. 2 variants

"a x 1 y b" b e {(2,3,4,5} - {x,y,a}. 1 variants

Therefore, N=12*2=24

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

27 Sep 2009, 10:48

At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number of different possible seating arrangements for the group?A. 5B. 10C. 24D. 32E. 120Soln: Since the arrangement is circular and 2 seating arrangements are considered different only when the positions of the people are different relative to each other, we can find the total number of possible seating arrangements, by fixing one person's position and arranging the others.Thus if one person's position is fixed, the others can be arranged in 4! ways.

Ans is C.

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

05 Nov 2010, 20:11

Bunuel wrote:

mybudgie wrote:

At a party, 5 people are to be seated around a circular table; two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements?A 5B 10C 24D 32

E 120

This is the case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\).The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:\(R = \frac{n!}{n} = (n-1)!\)"\((n-1)!=(5-1)!=24\)Answer: C.

Similar question: combinatrics-86547.html?hilit=relative%20around

Hope it's clear.

I was confused by the wording of the question "only when the positions of the people are different relative to each other". I knew the formula (n-1)!, but I though that the correct answer would require some limited set of the combinations, defined by "only when the positions of the people are different relative to each other".

Then if the question did not mentioned this special condition of ""only when the positions of the people are different relative to each other", the answer would be 24x5=120. Right?

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

05 Nov 2010, 20:40

Fijisurf wrote:

Bunuel wrote:

mybudgie wrote:

E 120

Similar question: combinatrics-86547.html?hilit=relative%20around

Hope it's clear.

Then if the question did not mentioned this special condition of ""only when the positions of the people are different relative to each other", the answer would be 24x5=120. Right?

"the positions of the people are different relative to each other" just means different arrangements (around a circular table). The number of arrangements of n distinct objects in a circle is \((n-1)!=4!=24\), (120 would be the answer if they were arranged in a row). _________________

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

06 Nov 2010, 04:08

Arranging 3 people (A, B, C) in a row:A B C, A C B, B A C. B C A, C A B, C B A3! waysWhy is arranging 3 people in a circle different? ....A ....O B......CIf I am B, A is to my left, C is to my right.Look at this one now:....C....OA......BHere also, A is to my left and C is to my right. In a circle, these are considered a single arrangement because relative to each other, people are still sitting in the same position. This is the general rule in circular arrangement. You use the formula n!/n = (n - 1)! because every n arrangements are considered a single arrangement. e.g. if n = 3, the given 3 arrangements are the same: .....A ................ C ............... B .....O ................ O .............. O B........C ........ A ..... B ..... C........ AIn each of these, if I am B, I am sitting in the same position relative to others. A is to my left and C is to my right.and these three are the same: .....C ................ A ............... B .....O ................ O .............. O B........A ........ C ..... B ..... A........ CHere, if I am B, C is to my left and A is to my right. Different from the first three.Hence no. of arrangements = 3!/3 = 2 onlyHere, they have mentioned 'relative to people' only to make it clearer. In a circle, anyway only relative to people arrangements are considered.You might need to use n! in a circle if they mention that each seat in the circular arrangement is numbered and is hence different etc. Then there are just n distinct seats and n people. If nothing of the sorts is mentioned, you always use the (n - 1)! formula for circular arrangement. _________________

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

23 Mar 2013, 21:00

Hi there,

You can treat this as an ordering question except that for a circular arrangement you need to divide by the number of spaces. So in this case:

5!/5=24If you spin the circle to right, that doesn't count as a new arrangement. Dividing by the number of spaces takes that into consideration.Happy Studies,HG. _________________

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At a dinner party, 5 people are to be seated around a circular table. [#permalink]

Updated on: 11 Oct 2022, 01:25

Val1986 wrote:

E. 120

Please check my signature for the link to the relevant blog posts. _________________

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Originally posted by KarishmaB on 24 Mar 2013, 21:01.
Last edited by KarishmaB on 11 Oct 2022, 01:25, edited 1 time in total.

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

23 Apr 2014, 18:41

VeritasPrepKarishma wrote:

Val1986 wrote:

E. 120

Check out this post on circular arrangements. It discusses why the number of arrangements is n!/n (which is the same as (n-1)!) in case there are n people sitting around a round table.

http://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html?/2011/10 ... angements/

It also discusses the relevance of this statement in the question: "Two sitting arrangements are considered different only when the positions of the people are different relative to each other"

Hi Karishma,

If there were constraints such as A can't be next to B or C, does that mean that we now have 5 seats but since 3 of them are fixed, the solution would be 2!/2?

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At a dinner party, 5 people are to be seated around a circular table. [#permalink]

Updated on: 16 Oct 2022, 23:39

russ9 wrote:

Hi Karishma,

If there were constraints such as A can't be next to B or C, does that mean that we now have 5 seats but since 3 of them are fixed, the solution would be 2!/2?

I am assuming your question is this:5 people are to be seated around a circular table such that A sits neither next to B nor next to C. How many arrangements are possible?I don't know how you consider "...3 of them are fixed". The way you handle this constraint would be this:There are 5 vacant seats. Make A occupy 1 seat in 1 way (because all seats are same before anybody sits).Now we have 4 unique vacant seats (unique with respect to A) and 4 people. B and C cannot sit next to A so D and E occupy the seats right next to A on either side. This can be done in 2! ways: D A E or E A DB and C occupy the two unique seats away from A. This can be done in 2! ways.Total number of arrangements = 2! * 2! = 4 _________________

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Originally posted by KarishmaB on 23 Apr 2014, 19:18.
Last edited by KarishmaB on 16 Oct 2022, 23:39, edited 1 time in total.

Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

22 Feb 2015, 09:43

Please provide feedback to see if this makes sense:I approached it thinking, if you have a circle, say, ABCDE there would be 5! ways of arranging, however, the question states that an arrangement is only different if the positions relative to each other are different. So (1/5)th of the time each person in the circle would in the same position relative to another person. Therefore I did (1/5)x5! = 24

I was thinking "symmetry" as well - what are the experts thoughts on this? The formula though is definitely the easier way.

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

22 Feb 2015, 14:00

Hi icetray,Your thinking on this question is just fine. Conceptually, since we're dealing with a circular table with 5 chairs (and not a row of chairs), the table could have 5 different "starting chairs." As such the arrangements (going around the table):ABCDEBCDEACDEABDEABCEABCDAre all the same arrangement (just 'revolved' around the table). Since we're NOT allowed to count each of those (they're not different arrangements, they're just rotations of the same arrangement), we have to divide the permutation by 5.5!/5 = 24This type of 'set-up' is relatively rare on Test Day - there's a pretty good chance that you won't see it at all. If you do though, then your way of handling the "math" is just as viable as the formula that was given.GMAT assassins aren't born, they're made,Rich _________________

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

01 Mar 2018, 17:37

Val1986 wrote:

E. 120

When determining the number way to arrange a group around a circle, we subtract 1 from the total and set it to a factorial. Thus, the total number of possible sitting arrangements for 5 people around a circular table is 4! = 24.Answer: C _________________

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

18 Dec 2019, 07:22

Hi, I'm not so sure about the (n-1)! way. Since the question stated that "Two seating arrangements are considered different only when the positions of the people are different relative to each other", wasn't this formula still count some invalid possibilities, for example: if we have 5 people, namely A,B,C,D,E. The arrangements of ABCDE or ACBDE will be counted as 2 different arrangements by the (n-1)! formula right? but the positions of D relatively to E or A in these cases are not different.

Am I wrong somewhere?

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

09 Mar 2020, 02:08

marcodonzelli wrote:

E. 120

total possible ways in circular table ( n-1)! ; (5-1)! ; 24

IMO C

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Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink]

24 May 2020, 15:26

Bunuel: really appreciate the "Questions about this concept to practice:" section! Super helpful to solidify concepts!

Wish all Qs have this... For other questions that don't, is there a quick way I can dig out similar questions-as you did here?

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