This is a question about permutations. A permutation is an arrangement of a certain amount of objects in a specific order. For example, if we have two people, #A# and #B#, we can arrange them in a row of two chairs in two ways #A B# or #B A# Each of these two arrangements is a permutation. Thus, there are only two possible permutations in the example. If we have seven people it gets a bit trickier. We must use the multiplication principle. If we have seven people and we want to know how many ways we can arrange them in a row of seven chairs, using the multiplication principle, we multiply all the options together to get the total number of arrangements. E.g. there are 7 options for the first seat, 6 for the second (because one has been used), 5 for the third, 4 for the fourth, and so on. #7*6*5*4*3*2*1=7! =5040# So there are 5040 ways of arranging seven people in a row of seven chairs. I think the proof of this is best seen by making a tree diagram and noting that the number of branches at the end tells you the number of possible arrangements/permutations. The tree for seven people is way too big so if you want to test this, do it with a smaller number, such as three people (there will be #3*2*1=6# branches/arrangements). I won't go into more detail on how this works, other than to say that it is very important and should be understood if you want to deal with permutations or probability. Now, let's call each person #A, B, C, D, E, F, G# The answer is not 5040 because we have a restriction, two of the people must sit next to each other. We'll call Jane and Joe, #A# and #B#. Group them together and call them one object, #X#. Now we effectively have six objects #X, C, D, E, F, G# where #X=A, B# Using the multiplication principle, we can arrange six objects in 6! ways #6*5*4*3*2*1=6! =720# But we also have to take into consideration the group, #X#, which can also be arranged in multiple ways #AB# or #BA# #2*1=2! =2# To take this into account and get the total number of arrangements, we must use the multiplication principle again. This involves multiplying the number of arrangements of six objects by the number of arrangements of the group, #X# #720*2=1440# This makes sense because we have 720 arrangements where #A# is first, but we can have another 720 arrangements where #B# is first.
 Senior Manager Joined: 23 Apr 2010 Posts: 487
7 people sit at the round table. In how many ways can we [#permalink] 7 people sit at the round table. In how many ways can we sit them so that Bob and Bill don't sit opposing each other?My solution:
Math Expert Joined: 02 Sep 2009 Posts: 87761
Re: 7 people round table [#permalink]
nonameee wrote: So, I have found the exact wording of the problem and I've made a mistake: Seven people are to be seated at a round table. Bill and Bob don't want to sit next to each other. How many seating arrangements are possible? So in that case, the correct solution is:6! - 2!*5!As for my original question (i.e., they Bob and Bill refuse to sit opposite each other, the correct solution, IMO, should be: 6! - 5!But I would still like some math experts to check it. Bunuel? Question #1: Seven people are to be seated at a round table. Bill and Bob don't want to sit next to each other. How many seating arrangements are possible? Total # of arrangements around the circular table: \((7-1)!=6!\);Arrangements when Bill and Bob sit next to each other: 5!*2: consider Bill and Bob as one unit - {Bill, Bob}. Now we have total 6 units: {Bill, Bob}, {1}, {2}, {3}, {4}, {5}. Total # of arrangements of these 6 units around the table is \((6-1)!=5!\) and Bill and Bob within the unit can be arranged in 2 way: {Bill, Bob} and {Bob, Bill};So, # of arrangements when Bill and Bob don't sit next to each other is: \(6!-2*5!\).Question #2: Seven people are to be seated at a round table. Bill and Bob don't want to sit opposite to each other. How many seating arrangements are possible? Now, if we consider "opposite" to mean "directly opposite", meaning that Bill and Bob shouldn't sit at the endpoints of the diameter of the circular table, then total # of arrangements will simply be \((7-1)!=6!\), because if 7 (odd) people are distributed evenly around a circle no two people are directly opposite each other (no diagonal in 7 sided regular polygon passes through the center).If the question were: SIX people are to be seated at a round table. Bill and Bob don't want to sit opposite to each other. How many seating arrangements are possible?Total # of arrangements around the circular table: \((6-1)!=5!\);Now, Bill can have 5 people opposite him. In 1/5 of the cases he'll be opposite Bob (in equal # of arrangements he'll be opposite each of these 5 people), so all but these 1/5 of the cases are acceptable (4/5 of the cases are acceptable). So # of arrangements when Bill and Bob don't sit opposite each other is: \(5!*\frac{4}{5}\). _________________
Retired Moderator Joined: 20 Dec 2010 Posts: 1224
Re: 7 people round table [#permalink]
nonameee wrote: Can someone please explain why we have to multiply 5! by 2? Thanks. Let's name them; 7 People: Bob,Bill,Red,Blue,White,Pink,PurpleRound Table: T7 Chairs: 1,2,3,4,5,6,7Let chairs 1 and 4 be opposite each other; Bob is sitting on 1 and Bill is sitting on 4; The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysNow, Bill and Bob swap their positions; Bob is sitting on 4 and Bill is sitting on 1;The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysThus total number of ways in which Bob and Bill are sitting opposite each other = 2*5! = 120*2 = 240We have to find in how many ways they are NOT sitting opposite; subtract 240 from the total number of possible arrangements;In circular permutation, Total number of arrangements of n people = (n-1)!Here number of people = 7Arrangements = (7-1)! = 6!= 720Number of ways Bob and Bill are NOT sitting opposite = 720-240 = 480.
Retired Moderator Joined: 20 Dec 2010 Posts: 1224
Re: 7 people round table [#permalink] Total number of ways in which all 7 people can sit; (7-1)! = 6!Let's fix Bob and Bill's positions to be opposite each other. Rest 5 can sit in 5! ways. Now, BOB and Bill toggle their position; rest 5 will rearrange themselves in 5! way again.Bob and Bill don't sit at the opposite tables = 6!-2*5! = 5!(6-2) = 4*5! = 4*120 = 480. What's the official answer?
Senior Manager Joined: 23 Apr 2010 Posts: 487
Re: 7 people round table [#permalink] 480 is the official answer. Why did you multiply 5! by 2 (2*5!)?
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Re: 7 people round table [#permalink] Consider Bob and Bill sitting next to each other. The order can be Bob to the right or left of Bill. This happens for all 5! ways the remaining 5 ppl arrange themselves in 5 places. Hence 5! for the 5 ppl multiplied by 2 ways in which Bob and Bill can sit next to each other
Senior Manager Joined: 23 Apr 2010 Posts: 487
Re: 7 people round table [#permalink] Can someone please explain why we have to multiply 5! by 2? Thanks.The reason I'm asking is this:When we have 4 couples and want to sit them so that a husband sits opposite his wife, we get:1*1*6*1*4*1*2*1 = 48 and we don't multiply it by 2. I am confused as these two questions seem pretty much the same to me.
Senior Manager Joined: 23 Apr 2010 Posts: 487
Re: 7 people round table [#permalink] OK, fluke, I understand everything. But can you please explain the difference between this question and the following one:When we have 4 couples and want to sit them so that a husband sits opposite his wife, we get:1*1*6*1*4*1*2*1 = 48 and we don't multiply it by 2. I am confused as these two questions seem pretty much the same to me.
Retired Moderator Joined: 20 Dec 2010 Posts: 1224
Re: 7 people round table [#permalink]
nonameee wrote: OK, fluke, I understand everything. But can you please explain the difference between this question and the following one:When we have 4 couples and want to sit them so that a husband sits opposite his wife, we get:1*1*6*1*4*1*2*1 = 48 and we don't multiply it by 2. I am confused as these two questions seem pretty much the same to me. Your reasoning seems correct, mine flawed.It should be 6!-5!=720-120=600 as reversing positions for Bill and Bob doesn't change relative positions for the rest of them.One such position where two positions will be considered same are; Bill,P1,P2,Bob,P3,P4,P5Bob,P3,P4,P5,Bill,P1,P2I too need an expert opinion on this one. Someone please confirm a good way to analyze this one.
Senior Manager Joined: 23 Apr 2010 Posts: 487
Re: 7 people round table [#permalink] So, I have found the exact wording of the problem and I've made a mistake: Seven people are to be seated at a round table. Bill and Bob don't want to sit next to each other. How many seating arrangements are possible? So in that case, the correct solution is:6! - 2!*5!As for my original question (i.e., they Bob and Bill refuse to sit opposite each other, the correct solution, IMO, should be: 6! - 5!But I would still like some math experts to check it. Bunuel?
Senior Manager Joined: 23 Apr 2010 Posts: 487
Re: 7 people round table [#permalink] Bunuel, thanks for your solutions. Great as always.
Manager Joined: 21 Mar 2010 Posts: 242
Re: 7 people round table [#permalink] We dont know how many people the table can seat. if it can seat 10/20 etc etc. If you have 7 people sitting on a table that can seat 20, then if there can be no empty spots between people, the probability that any two people will sit opposite each other is 0? Also how do you seat 7 people on a round table evenly?
Math Expert Joined: 02 Sep 2009 Posts: 87761
7 people sit at the round table. In how many ways can we [#permalink]
mbafall2011 wrote: We dont know how many people the table can seat. if it can seat 10/20 etc etc. If you have 7 people sitting on a table that can seat 20, then if there can be no empty spots between people, the probability that any two people will sit opposite each other is 0? Also how do you seat 7 people on a round table evenly? This does not make ANY sense. The question is supposed to be a PS problem and if we cannot safely assume that there are exactly 7 seats then it has no solution at all.Next, you can distribute 7 people evenly around the table the same way as 2, 3, ..., or 100 people: divide the circumference into 7 equal parts, mark it and place a person there. _________________
Manager Joined: 19 Apr 2011 Posts: 63
Re: 7 people round table [#permalink] Very good question
Re: 7 people sit at the round table. In how many ways can we [#permalink] Can we not solve this in a way where we don't subtract the "opposite" case from the total possibilities?Bob can select the seats in 1 way since this is a circular table. Bill can sit on any seat apart from the opposite seat so 6 remaining seats - 1 = 5 ways and rest of the people can arrange themselves in 5! ways.So 5 x 5! way = 5x5x4x3x2 = 600. I know this isn't the answer. Can anyone explain why is my approach wrong?
Intern Joined: 18 May 2016 Posts: 24
Re: 7 people sit at the round table. In how many ways can we [#permalink] Another way to see the answer for opposite, again assuming we only care about relative positioning. I take the example where there are 6 guests. Bob has 6 slots to choose from. Bill has then 5-1=4 slots. The rest have 4!. So answer is 6*4*4!, which we divide by 6 for the number of potential rotations around the table.
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Re: 7 people sit at the round table. In how many ways can we [#permalink] 6!-2×5!=720-240=480 Posted from my mobile device
Re: 7 people sit at the round table. In how many ways can we [#permalink]
fluke wrote: nonameee wrote: Can someone please explain why we have to multiply 5! by 2? Thanks. Let's name them; 7 People: Bob,Bill,Red,Blue,White,Pink,PurpleRound Table: T7 Chairs: 1,2,3,4,5,6,7Let chairs 1 and 4 be opposite each other; Bob is sitting on 1 and Bill is sitting on 4; The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysNow, Bill and Bob swap their positions; Bob is sitting on 4 and Bill is sitting on 1;The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysThus total number of ways in which Bob and Bill are sitting opposite each other = 2*5! = 120*2 = 240We have to find in how many ways they are NOT sitting opposite; subtract 240 from the total number of possible arrangements;In circular permutation, Total number of arrangements of n people = (n-1)!Here number of people = 7Arrangements = (7-1)! = 6!= 720Number of ways Bob and Bill are NOT sitting opposite = 720-240 = 480. are you saying that in a 7 seater round table only 2 chairs are opposite each other?
7 people sit at the round table. In how many ways can we [#permalink] fluke Quote: Let chairs 1 and 4 be opposite each other;Bob is sitting on 1 and Bill is sitting on 4;The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysNow, Bill and Bob swap their positions;Bob is sitting on 4 and Bill is sitting on 1;The rest of the people on 2,3,5,6,7 - can rearrange themselves in 5! waysThus total number of ways in which Bob and Bill are sitting opposite each other = 2*5! = 120*2 = 240We have to find in how many ways they are NOT sitting opposite; subtract 240 from the total number of possible arrangements;In circular permutation, Total number of arrangements of n people = (n-1)!Here number of people = 7Arrangements = (7-1)! = 6!= 720 Number of ways Bob and Bill are NOT sitting opposite = 720-240 = 480 Even if we consider 1 & 4 opposite each other, the remaining arrangements will only be 5! ways. We do not need to multiply it again with 2 because even if we change the positions of 1 & 4 we will get the same arrangement wrt to each other. Bob-2-3-Bill-5-6-7 is same as Bill-3-2-Bob-7-6-5. In one case it will be clockwise and in other it will be anti-clockwise, but the respective positions will remain same.Kindly refer to image.So the answer will be 6!-5!
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Re: 7 people sit at the round table. In how many ways can we [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: 7 people sit at the round table. In how many ways can we [#permalink] 24 May 2022, 12:29 |
How many ways can 7 people can be seated at a round table?
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