In how many ways can 6 people be seated around a circular table if

Total ways to arrange 6 people in a circle = (6 - 1)! = 120

Total ways to arrange 6 people in a circle such that 2 people are always together = (5 - 1)! = 24 and these two people can be arranged in 2! ways. Therefore, total 24 * 2 = 48 ways

Total ways to arrange 6 people in a circle if 2 particular people are always separated: 120 - 48 = 72

Answer C

The number of circular permutations of n objects = (n-1)!

The total number of ways in which 6 people can be arranged around a circle is therefore 5! = 120, which is not our answer since this is not what we want. Answer option E can be eliminated.

Since we want to find the number of arrangements where 2 people are always separated, we can find the total arrangements (i.e. 120) and then subtract the number of arrangements where these two people are together.

Number of permutations where A and B (say) are together = Total permutations – Number of permutations where A and B are together

Number of permutations where A and B are together:
Since A and B have to be together, we consider them as ONE object. So, we now have a total of 5 objects (including the group of A&B). 5 objects can be arranged around a circle in 4! = 24 ways.

But, for each of these 24 ways, A and B can be arranged in 2! Ways.
Therefore, number of permutations where A and B are together = 4! * 2! = 48.

So, required number of permutations = 120 – 48 = 72.

The correct answer option is C.

In P&C questions with restrictions, using the backdoor approach saves time and effort and also ensures that we don’t get muddled in the various cases that the forward approach entails. Solving this question would be infinitely more difficult if we went about taking cases for organizing A and B separately.

Hope that helps!
Aravind B T

we have already taken out the relative position i.e AB & BA are out in 120, so it should not be subtracted
e.g. ABCDEF or BACDEF only one of them is counted in 120, but while subtracting you are using both case.
CrackVerbal

tramobile

You are considering the arrangement ABCDEF and BACDEF as a single arrangement and think that it is counted only once in the total 120 arrangements.

Let's analyze your above statement in detail.

Remember, 2 seating arrangements are considered different when the positions of the people are different relative to each other.

The example you have mentioned, i.e ABCDEF and BACDEF are 2 different arrangements because, in the arrangement ABCDEF, B is sitting between A and C in the circular table while in BACDEF, B will be sitting between F and A. Here, the position of B is different in both cases, Hence it should be counted as 2 different arrangements.

I hope this is clear for you and that's the reason we are subtracting both cases i.e 2!*4! from the total 120 arrangements.

Let's us know if you have any further queries.

Thanks,
Clifin Francis,
GMAT Mentor

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