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

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A group of 8 friends sit together in a circle. If A refuses to sit bes  [#permalink]

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New post 18 May 2019, 06:58
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Question Stats:

23% (03:00) correct 77% (02:56) wrong based on 31 sessions

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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

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Re: A group of 8 friends sit together in a circle. If A refuses to sit bes  [#permalink]

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New post 18 May 2019, 06:59
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dabaobao wrote:
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


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
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A group of 8 friends sit together in a circle. If A refuses to sit bes  [#permalink]

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New post 22 Oct 2019, 04:30
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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.
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A group of 8 friends sit together in a circle. If A refuses to sit bes   [#permalink] 22 Oct 2019, 04:30
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