A group of 8 friends sit together in a circle. If A refuses to sit bes

Question

A group of 8 friends sit together in a circle. If A refuses to sit beside B unless C sits on the other side of A as well, how many possible seating arrangements are possible?

A) 3600
B) 3840
C) 4440
D) 31,680
E) 36,000

axezcole · Accepted Answer

Another way -

I call it imagine people entering a meeting room one by one/ VeritasKarishma's method.
Refer- https://gmatclub.com/forum/combinatorics-made-easy-206266.html 
Topic name-circular arrangements

We have been said that A can sit besides B unless c sits on other side of A.
So our answer is ways in which A can sit besides B unless c sits on other side of A
+ other ways

Two conditions are possible-

A can sit besides B unless c sits on other side of A=1*2*1*5! (A comes in 1st and has 1 way to decide,B comes in 2nd and has 2 ways to decide,C comes in 3rd and has 1 way to decide,Rest have 5 ! ways to decide)

Other ways(remember B cannot be adjacent to A here)=1*5*6! (A comes in 1st and has 1 way to decide,B comes in 2nd and has 5 ways to decide, Rest have 6 ! ways to decide)

Final answer-
1*2*1*5! +1*5*6! =3840

Solve using a circular table diagram for the two conditions mentioned,it will make more sense.

dabaobao · Answer

Official Solution  
Credit: Veritas Prep

Solution: Let’s start with what we know. We know that the total number of ways in which 8 people can be arranged around a circular table is (8−1)!=7!(8−1)!=7!

Since we do not want A to sit next to B, let’s try and make them sit together. This will give us the number of arrangements that are unacceptable to us. Let’s say that A and B are a single unit. So now there are 7 units which need to be arranged in a circle. This can be done in (7−1)!=6!(7−1)!=6! ways. Since there are two arrangements possible, AB and BA, within the unit, we need to multiply 6! by 2.

Number of arrangements in which A and B sit together =2∗6!=2∗6!

We can subtract these ‘unacceptable arrangements’ from total arrangements to get the number of ‘acceptable arrangements’. But this number of ‘unacceptable arrangements’ includes those arrangements where C is sitting on the other side of A. But those arrangements are acceptable to us so we should not subtract them out. How many such arrangements are there in which A and B are sitting together and C is sitting beside A too?

Now C, A and B form a single unit leaving us with 6 units to be arranged in a circle. 6 units can be arranged in (6−1)!=5!(6−1)!=5! ways

CAB can also be arranged as BAC, hence the 5! needs to be multiplied by 2. (Mind you, we will not consider ABC, ACB etc here since A should be in the middle)

Number of arrangements in which A and B sit together and C sits beside A = 2∗5!2∗5!

Therefore, number of unacceptable arrangements =2∗6!–2∗5!=2∗6!–2∗5!

We subtract these out of the total number of arrangements and we get the total number of acceptable arrangements.

Possible number of seating arrangements = 7!–(2∗6!–2∗5!)=38407!–(2∗6!–2∗5!)=3840

If you are wondering about the ‘painful’ calculation involved in the step above, don’t worry. Calculations with factorials are generally quite straight forward.

7!–(2∗6!–2∗5!)=7!–2∗6!+2∗5!7!–(2∗6!–2∗5!)=7!–2∗6!+2∗5!

=2∗5!(21–6+1)=2∗5!(21–6+1) (Take 2*5! common out of the three terms)

=2∗120∗16=32∗120=3840

chetan2u · Answer

In such we should fix a position of some one and then count ways in which others can be seated with given conditions.

Since, the seating and restrictions related to it revolve around A, let us fix position of A.

Now, B can sit next to him in two ways but then C has to be seated in the other portion => 2 ways
Remaining 5 ( less A, B and C) can be seated anywhere on the remaining 5 seats => 5!
Total 2*5!

Apart from the above methods B cannot sit next to A, so B has 5 ways to sit.
The remaining 6 can be seated in any of the remaining 6 seats => 6!
Total 5*6!

Combined total 2*5!+5*6! = 5!(2+30)=120*32=3840

B

Mona2019 · Answer

The number of ways A and B can sit together with C not on A's side.
Considering AB as one unit, they can take any position . The place next to A can be filled in 5 ways (excluding C)
AB can be seated in 2 ways.
The rest of the chairs can be filled in 5! Ways.
So total number of ways = 2*5*5! = 1200

Total number of ways 8 people can be seated = 7!

So required number of ways = 7!- 1200 = 3840

Posted from my mobile device

pierjoejoe · Answer

the point here is that we have 8 friends. 
total possible combinations (8-1)! = 7!
we have to remove all the arrangements where A sits next to B and C is NOT on the other side

we have 2 situations we want to avoid, B seats to the right of A and B seats to the left of A.
each one have 6! combinations of the other 6 people
AB --> 6! (B to the right of A)
BA --> 6! (B to the left of A
total = 2*6!

but in these combinations we have some acceptable arrangements where A sits next to B and on the other side there is C
each of these combinations have 5! possible arrangements of the other 5 people
CAB --> 5!
BAC --> 5!
total = 2*5!

we have to remove these arrangements from 2*6! 
2*6! - 2*5! 
now the resulting number is the number of ways A sits next to B (AB or BA) and C DOES NOT sit on the other side.

we can subtract from the total of 7! possible combination this number:

7! - (2*6! - 2*5!) = 3840

bumpbot · Answer

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A group of 8 friends sit together in a circle. If A refuses to sit bes

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