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# In how many ways can 6 people be seated at a round table if

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Joined: 02 Sep 2009
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In how many ways can 6 people be seated at a round table if  [#permalink]

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22 Jun 2016, 02:36
1
asicconi wrote:
So is the correct answer 36? The original post says 108.

The question is a bit ambiguous. Here is the solution to why the answer is 108: in-how-many-ways-can-6-people-be-seated-at-a-round-table-if-36750.html#p253783

Check the discussion HERE for more.
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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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28 Jul 2016, 03:39
sandeepmanocha wrote:
6 People Sitting Around a Round Table Without Any Restriction = (6-1)! = 5! = 120

Restriction = 1 person cannot sit around other two particular people
Complement Condition = 3 People Will Always Sit together
Now considering 3 People as one group along with other 3 people , total number of ways they can sit = (4-1) = 3! = 6 Ways
But group of 3 Can also Adjust it self in 3! ways = 6 Ways
Total Complement Ways = 6+6 = 12

Total Ways = 120 - 12 = 108

Please correct, if this is not the right way of solving the question

With all due respect sir
According to me every thing is right about your attempt...as I have also used somewhat the same approach
The only difference was that rather than performing 6+6, I performed 6*6..and hence my answer was 120-36=84.
The reason behind this multiplication is as follows
Both the steps, i.e. making a group of 3 people and then arranging them circularly & then arranging the 3 persons which are in a group they all have to be performed together to achieve the given result...we use addition when both choices alone are suuficient to give us the answer....
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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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28 Jul 2016, 03:47
AK47 wrote:
n people can be seated around a round table in (n-1)! ways.

ok...let's find the no of ways in which that person is always seated next to 2 particular people.these 3 can be seated in 2 ways because the cetre position is fixed.
now we have a total of 3+1 people...note that 1 represents the group of those 3 people.
so 4 can be seated in (4-1)! ways = 6 ways.
hence total ways when 2 particular people are always next to one particular of them = 2*6=12 ways..

and total no of ways in which 6 people can be seated =(6-1)!=120 ways..

hence answer= 120=12 = 108 ways.

choice b as per me.
what's the OA?

londonluddite wrote:
In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to 2 of the other 5?

a) 720
b) 120
c) 108
d) 84
e) 48

With all due respect sir
Can you please let me know what is wrong in my attempt
6 People Sitting Around a Round Table Without Any Restriction = (6-1)! = 5! = 120
Restriction = 1 person cannot sit around other two particular people
Complement Condition = 3 People Will Always Sit together
Now considering 3 People as one group along with other 3 people , total number of ways they can sit = (4-1) = 3! = 6 Ways
But group of 3 Can also Adjust it self in 3! ways = 6 Ways
hence total number of arrangements in the complement position=6*6
Both the steps, i.e. making a group of 3 people and then arranging them circularly & then arranging the 3 persons which are in a group they all have to be performed together to achieve the given result..
120-36=84.

Please let me know my mistake????
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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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30 Jul 2016, 09:58
bgpower wrote:
I knew the (n-1)! formula which doesn't solve the entire problem here so I shifted gears and came to the following approach.

You have 6 spots on the table.
Let's imagine to fix the guy who can't sit with all the other ones on one of the spots.
(1) As he can't stand 2 out of total of 5 people, we have 3 options for the 2 seats next to him - so 3C2 which equals 3.
(2) For the remaining 3 spots we have 3! or 6 options.
Multiply (1) and (2) and you get 18 seating arrangements.

But bear with me, we have to remember that we fixed the bad guy on just one spot. He can sit on every seat on the table, namely - 6.
So multiply the 18 seating arrangements with 6 and you get 108 (C)

Quote:
Very well explained, thank you!

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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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13 Aug 2016, 11:56
1
londonluddite wrote:
In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to 2 of the other 5?

a) 720
b) 120
c) 108
d) 84
e) 48

This kind of language for students who are in a learning phase is a crime against learning. This is so confusing.
The question should have stated that 1 person cannot sit between 2 people. Sitting next does not mean that the person is in between.
Example of sentences: I sat next to my nephew and niece who are sitting together. I am sitting next to a couple of bushes in the garden which are adjacent to each other. My house is next to the two red painted houses on the street.

If the person does not want to sit in between two specific people.
Answer = Total number of cases - Number of cases when he is sitting between these 2 people

Total = (6-1)! [number of ways of making people sit in a circle] = 120
Number of cases when the person is sitting between the 2 specific people. Place the 3 people aside a bind them together, and now we will consider them as one person. Now, we have 3 free people + 1 hypothetical person (which is actually those 3 binded) = Total 4
Formula for circular combination (n-1)! = (4-1)! = 3!
Now the three which we binded, can also be arranged.
The person which has issues with the other 2 will be in between and the other 2 can only be arranged in 2 ways next to him.
So the total ways in which the guy with issues will be stuck between the people he does not want to is 3!*2=12.

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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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28 Feb 2018, 13:48
Let Persons be A,B,C,D,E,F

Let Person B, cannot sit in between A and C.

We will use inversion technique to find the required answer.
N(B cannot sit between A and C) = total combinations - N(B will sit in between A and C)
total combinations = (6-1)! = 120

now to find out N(B will sit in between A and C), we will consider ABC as single person K
now total circular combinations (K,D,E,F) = (4-1)! = 6
now K will be {ABC, CBA}, so we have multiply by 2
so N(B will sit in between A and C) = 6 * 2 = 12

so final required N(B cannot sit between A and C) = total combinations - N(B will sit in between A and C)
= 120 - 12 = 108
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Re: In how many ways can 6 people be seated at a round table if  [#permalink]

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08 Apr 2019, 11:07
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Re: In how many ways can 6 people be seated at a round table if   [#permalink] 08 Apr 2019, 11:07

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