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 PrepSolution: 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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