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7 people sit at the round table. In how many ways can we

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7 people sit at the round table. In how many ways can we  [#permalink]

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New post 08 Feb 2011, 04:05
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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:

We solve the reverse problem: In how many ways can we sit them so that Bob and Bill sit opposing each other?

positive cases = 1*1*5!
all cases = 1*6!

don't sit opposing each other = 6! - 5!
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 06:43
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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: \(5!*\frac{4}{5}\).
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 01:26
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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,Purple
Round Table: T
7 Chairs: 1,2,3,4,5,6,7

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! ways

Now, 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! ways

Thus total number of ways in which Bob and Bill are sitting opposite each other = 2*5! = 120*2 = 240

We 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 = 7
Arrangements = (7-1)! = 6!= 720

Number of ways Bob and Bill are NOT sitting opposite = 720-240 = 480.
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Re: 7 people round table  [#permalink]

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New post 08 Feb 2011, 04:28
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?
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Re: 7 people round table  [#permalink]

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New post 08 Feb 2011, 04:38
480 is the official answer. Why did you multiply 5! by 2 (2*5!)?
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Re: 7 people round table  [#permalink]

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New post 08 Feb 2011, 04:54
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
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 01:06
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.
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 01:31
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.
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 03:23
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,P5
Bob,P3,P4,P5,Bill,P1,P2

I too need an expert opinion on this one. Someone please confirm a good way to analyze this one.
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Re: 7 people round table  [#permalink]

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New post 09 Feb 2011, 04:02
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?
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Re: 7 people round table  [#permalink]

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New post 10 Feb 2011, 00:16
Bunuel, thanks for your solutions. Great as always.
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Re: 7 people round table  [#permalink]

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New post 10 Feb 2011, 19:10
Bunuel
Im confused with your approach of the the second part where they dont want to sit opposite each other. We dont know how many seats are on the table and we dont know if there can be an empty spot on the table.

So how did you approach this qn?
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Re: 7 people round table  [#permalink]

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New post 11 Feb 2011, 04:23
mbafall2011 wrote:
Bunuel
Im confused with your approach of the the second part where they dont want to sit opposite each other. We dont know how many seats are on the table and we dont know if there can be an empty spot on the table.

So how did you approach this qn?


What empty spots? ...

Anyway: assume they are evenly distributed around the table.
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Re: 7 people round table  [#permalink]

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New post 11 Feb 2011, 08:33
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?
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Re: 7 people round table  [#permalink]

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New post 11 Feb 2011, 08:49
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 can not 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.
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Re: 7 people round table  [#permalink]

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New post 15 Jun 2011, 05:57
Very good question
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Re: 7 people sit at the round table. In how many ways can we  [#permalink]

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New post 25 Jan 2017, 21:00
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?
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Re: 7 people sit at the round table. In how many ways can we  [#permalink]

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New post 21 May 2018, 06:25
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 &nbs [#permalink] 21 May 2018, 06:25
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