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

Question

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

AK47 · Accepted Answer

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?

bgpower · Answer

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)

anindyat · Answer

6 People in a round table can be seated in (6 - 1) ! ways = 120.

Now we need to subtract the number of cases when one of those is sitting next to 2 of the other 5.

We can consider as if 5 people are sitting in a row because it is round table. 
Again consider 3 people, those who can not sit together, as a single unit â€“ 

So the possible arrangements among remaining people 5 â€“ 3 + 1 Unit are = 3 !
And the 3 people unit can arrange among themselves in 3 ! ways.

So the possible cases when one of those is sitting next to 2 of the other 5 = 3 ! * 3 ! = 36

Total possible cases = 120 -36 = 84

londonluddite · Answer

OA is C.

OE: 6 people can be seated round around a table in 5! ways (would appreciate someones clarification on whether this is correct and why). There are 2 ways the two unwelcome guests could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from 5! giving a result of 108.

Clear as mud :-D

Edit : AK why can n people be seated in (n-1)! ways and not n!

Thanks

Edit : AK why can n people be seated in (n-1)! ways and not n! Thanks

Bunuel · Answer

Check other  Seating Arrangements in a Row and around a Table Questions .

GMATinsight · Answer

Have Modified the Language to make it clearer

6 people are A, B, C, D, E and F
and B can not sit next to A and C

Considering the Position of B is fixed,

We have to make sure that 2 person who sit next to B are out of D, E and F

i.e. No. of ways of choosing the neighbours of B = 3 C 2 = 3

and the no. of ways the Selected neighbours can arrange at the two position adjacent to B = 2!

i.e. The ways the B and The neighbours can be arranged = 3 C 2 *2! = 3*2 = 6

Now The no. of ways in which Remaining Three Individuals can be arranged on remaining 3 seats = 3!

Total Ways of making Six person seated such that B doesn't sit next to A and C = 3 C 2 *2!*3! = 36

sunita123 · Answer

Hi,
Can i solve this by taking total number of arrangements i.e 5!=120 ways and then subtract the restriction?

120-(arrangements B should not sit next to A and/or C) ?

is that a correct approach?

GMATinsight · Answer

That would be fine but a difficult approach as you will have to calculate three cases

Case-1 : When A sits next to B and C does not sit next to B
A can sit next to B in 2 ways (On B's right or B's left side)
The next adjacent place of B can be occupied in 3 ways because C can't sit next to B 
Remaining three can sit in 3! ways
So total ways = 2*3*3! = 36 ways

Case-2 : When C sits next to B and A does not sit next to B
C can sit next to B in 2 ways (On B's right or B's left side)
The next adjacent place of B can be occupied in 3 ways because A can't sit next to B 
Remaining three can sit in 3! ways
So total ways = 2*3*3! = 36 ways

Case-3 : When A and C both sit on either sides of B
A and C can sit in 2! ways on two places adjacent to B
Remaining three can sit in 3! ways
So total ways = 2!*3! = 12 ways

Total Unfavourable cases = 36+36+12 = 84 ways

Total favourable Cases = (6-1)! - 84 = 120 - 84 = 36 ways

I hope it helps!

sandeepmanocha · Answer

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

Bunuel · Answer

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.

letlovebethejudge · Answer

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)!   = 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.

Answer now is 120-12=108.

Manjotkaur77 · Answer

The total number of ways of arranging 6 people around a circular table = (6-1)! = 5!=120

Assuming A can't sit next to B & C.

Going with the complement condition, we assume those cases in which A sits next to B & C. Taking these 3 people as one unit, the total number of ways of arranging the remaining 3 people and this 1 unit (= 3+1) around a circular table is 3!. Now B & C can swap their positions around A as well. So, number of ways are= 3!*2=12.

Now, subtracting the complement condition from the total number of possible arrangements, we get our required answer, => 120-12=108.

Dooperman · Answer

Supposing A can’t sit next to B and C

When considering A,B,C together, 3! arrangements are being considered. 
But this also includes the arrangement ABC where A is sitting only adjacent to B and not to C.

the correct answer should not be 108. Kindly explain further.

ritik0212 · Answer

Hi,

Is it 108 or 36? depends on what cases needs to be considered as per the question.

1. Case 1 : one individual can’t sit next to 2 of these people. (Taken both at a time). So cases where this is going to happen are (2 ways of placing these individuals around our choosey individual) * 3! Ways of placing the remaining ones on table = 12 cases.
Eliminating these 12 from a total of (6-1)! Gives us  108.

2. Case 2 : one individual can’t sit next to 2 of these people. (Taken both at a time + next to either of them). So here along with the last 12 cases. There are cases when either one of these 2 individuals sit net to our choosey person.

There are 2 ways to calculate this. 
A) Straight away place choosey one (1) * place any of the others except those 2 around choosey one (3*2). Place rest of the 3 in the 3 positions 3! —> 6*6 =  36

B) 2nd way of doing it would be removing the (either 1 of these) cases along with the first 12 from 108. That’s a lengthy method but you can try solving it that way to get more clarity on the cases in this question.

Regards,
Ritik

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

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