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

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 –

So the possible arrangements among remaining people 5 – 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

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

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

Thanks

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

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 = 3C2 = 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 = 3C2 *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 = 3C2 *2!*3! = 36

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?

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!

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

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.

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.

Answer now is 120-12=108.

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.

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