Six people are to be seated at a round table with seats arranged at eq

nick1816
Can anyone please confirm if this method is correct.
Among 6 people sitting in a circle, Bob can sit on 6 places*Andy can sit on 4 places(as one place is taken by Bob and another one is opposite, so 4)*rest 4 people

6*4*4 = 96.

I doubt this approach is correct, because rest of the 4 people will have 4! Ways to sit. Also in the first place wherever Andy it's the same choice so there is only one choice for Andy to sit. So one choice for Andy and four choices for Bob and remaining 4 people 24 choices so answer is 4×24=96

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Bob can sit on 6 places*Andy can sit on 4 places(as one place is taken by Bob and another one is opposite, so 4).

You were correct upto this point. The remaining 4 people can be arranged in 4! ways. Since the positions or the chairs are indistinguishable (not marked), we have to consider only those permutations in which their(people) relative order changes. Hence, you have to divide by 6. \((\frac{6*4*4!}{6})\). You will get 4*4! = 96.

Is there a way to arrive at the solution using the Total Arrangements - Disallowed Arrangements method?

Total arrangements = 5! = 120
Ways in which Andy and Bob can sit opposite to each other = 2 * 4! (4! is the number of ways other 4 people can interchange seats)
I was multiplying by 2 since I was considering that Andy and Bob could interchange places too and still be opposite to each other.

So I ended up getting, disallowed arrangements = 2 * 4! = 48.
Therefore allowable arrangements = 120 - 48 = 72 which is incorrect.

VeritasKarishma, is it not required to consider Andy and Bob switching places between themselves? Since the arrangement of other 4 friends already accomplishes the different cases?

In that case, the number of disallowed arrangements will be 4! = 24, so number of allowed arrangements = 120 - 24 = 96 ways.
Thanks for your help !

Say we need Andy and Bob to sit opposite to each other.

Initially there are 6 identical seats so Andy sits anywhere. Now all seats are distinct. How many options are there for Bob to sit out of the rest 5? Only one seat. Switching will not give a different arrangement because relative to each other, they both will still be in the same positions.
The other 4 can sit in 4! ways. Hence, Andy and Bob can sit opposite to each other in 4! ways.

For a question to be a "circular permutation question", It is not enough for the things in the question to be arranged in a circle. After all, if you imagine a dining room with a circular table, with perhaps one chair by the door, you'd still consider one seating arrangement completely different from another if Ajike was by the door in the first arrangement, and Bill was by the door in the second, no matter how the people were arranged relative to each other. A question is only a circular permutation question if it explicitly tells you one arrangement is only different from another because of the relative positions of the things in the circle. This is how the actual GMAT will always phrase such a problem:

https://gmatclub.com/forum/at-a-dinner- ... 49709.html

In the particular problem posted above, assuming it is a genuine circular permutation question, then since we don't care about anything more than where people sit relative to each other, we can just fix person A in the 'first' seat. With no restrictions, we'd then have (5)(4)(3)(2)(1) = 120 choices for the remaining seats. We can then subtract the arrangements we're not allowed to use. If A and B did sit opposite each other, we'd then have 1 choice for the fourth seat, and then 4, 3, 2 and 1 choices for the remaining seats, for (4)(3)(1)(2)(1) = 24 arrangements that we can't use, leaving us with 120 - 24 = 96 that we can.

There are many ways in which the 2 guys do NOT sit directly opposite from each other, but only 1 instance where they are directly opposite to each other (since it is a circular permutation, each seat is identical at the beginning and each seat is only identifiable by its relative position to other people)


1st) Find the total number of ways to sit the 6 people with NO CONSTRAINTS

(N - 1)! = 5! = 120


2nd) SUBTRACT all the arrangements that have the 2 men sitting directly opposite from each other ——find the no of arrangements that VIOLATE the restriction


Rule: In arrangement problems, it’s usually best practice to start with the most constrained elements FIRST

1st) we seat Andy

Since in the beginning all the seats are identical, empty, and arranged in a circle, any seat that Andy chooses will be Identical to the next one.

Andy therefore only has 1 option.

Next, we need to sit Dan so as to violate the condition. We want them directly opposite from each other.

Only 1 seat will be directly opposite relative to Andy. Therefore, Dan will only have 1 option.

For the other 4 seats, now that we have 2 men seated, they are no longer identical and can be identified relative to the men seated. (To the left of Andy, to the right of Andy, etc.).

We can arrange the 4 ppl remaining in 4! Ways.


Total no of ways in which Andy and Dan do NOT sit directly opposite = (Total No. of Arrangements with NO Restriction) - (No. of Arrangements that VIOLATE the restriction and have Andy and Dan directly opposite each other)

(5!) - (1 * 1 * 4!) =

120 - 24 =

96 Arrangements

-E-


EDIT: Months later upon 2nd Inspection.....

Using the Concept of Symmetry the Question can be answered much more quickly.

(1st) Find the No. of Possible Arrangements Ignoring the Violation
5! = 120

(2nd) Since it is a Circular Permutation Problem, we only care about the Relative Position. The seats in and of themselves are not unique: only the placement of each person relative to the other.

For all of the 120 Arrangements, there is an equal probability that 1 of 5 scenarios will occur:

Sc. 1: Andy is directly across from Bob

Sc. 2: Andy is 2 Seats to the Left of Bob

Sc. 3: Andy is Directly to the left of Bob

Sc. 4: Andy is 2 Seats to the Right of Bob

Sc. 5: Andy is Directly to the Right of Bob

Out of the Entire 120 Possible Arrangements, the only Arrangements we want removed are Sc. 1 in which Andy is Directly Across from Bob.

This will occur in 1/5th of the 120 Arrangements:

120 - (1/5)*120 = 120 * (4/5) =

96

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