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There are 6 people at a party sitting at a round table with 6 seats: A

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Re: There are 6 people at a party sitting at a round table with 6 seats: A  [#permalink]

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New post 03 Jun 2019, 01:39
rishabhjain13 wrote:
VeritasKarishma wrote:
mrdanielkim wrote:
i figured it out after taking a break from studying. thanks anyway!

here's the solution anyway, though the book has the question worded differently:

Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:

A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108.


This question is different from the one posted above.

There are two versions and the answer would be different in the two cases.

Let me pick this version first:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side)

Total number of ways of arranging 6 people in a circle = 5! = 120

In how many of these 120 ways will A be between D and F?

We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways.
or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways.
In all, we make A sit next to D and F simultaneously in 12 ways.

120 - 12 = 108 is the number of ways in which D and F are not sitting next to A at the same time.


The second version which seemed like the intended meaning of the original poster:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.)

Here, we say that A has to sit next to two of B, C and E.
Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6

The rest of the 3 people can sit in 3! = 6 ways
Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways


Hi VeritasKarishma

I used another method learnt from one of your posts itself, to solve this question. Please see if the application is correct.

If A can sit neither next to D nor next to F, I calculated the total cases (5!) from which I subtracted all the cases where A and D would necessarily sit together (2*4!) and the ones where A and F would necessarily sit together (2*4!). Now, here, there would be a slight overlap where A would be sitting between D and F (DAF and FAD), so adding that to the equation to remove the overlap (2*3!)

5! - (2*4! + 2*4! - 2*3!)
5! - 2*4! - 2*4! + 2*3!
120 - 48 - 48 + 12
=36 ways


Yes, you are using the sets concept here to solve the question. It is perfectly correct.
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Re: There are 6 people at a party sitting at a round table with 6 seats: A  [#permalink]

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New post 03 Jun 2019, 05:48
Selecting 1st seat for A : No. of available seat = 6 No. of ways to select 1 seat for A = 6
Now as we have selected one seat we can no longer apply circular arrangement formula.
Now selecting seat for D and F : No. of available seat = 3 (As we can not place them on either side of A) No. Of ways to select 2 seat out of 3 for D and F = 3
Now arranging B C and E in remaining 3 seats = 3!
Total Nos of possible arrangements = 6*3*3! = 108

( Experts kindly review the approach is it correct?)
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There are 6 people at a party sitting at a round table with 6 seats: A  [#permalink]

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New post 22 Oct 2019, 04:07
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 neither next to D nor next to F.
So our answer is total ways- ways in which A can sit next to D or next to F.

Three conditions are possible-

A sits between D& F =1*2*1*3! (A comes in 1st and has 1 way to decide,D comes in 2nd and has 2 ways to decide,F comes in 3rd and has 1 way to decide,Rest have 3 ! ways to decide)

A sits only next to D(remember F cannot be adjacent to A here)=1*2*3*3! (A comes in 1st and has 1 way to decide,D comes in 2nd and has 2 ways to decide,F comes in 3rd and has 3 ways to decide,Rest have 3 ! ways to decide)

A sits only next to F(remember D cannot be adjacent to A)=1*2*3*3! (A comes in 1st and has 1 way to decide,F comes in 2nd and has 2 ways to decide,D comes in 3rd and has 3 ways to decide,Rest have 3 ! ways to decide)

Final answer-
1*2*3!+ 1*2*3*3!+ 1*2*3*3!=84


5!-84=36

Solve using a circular table diagram for the three conditions mentioned,it will make more sense.
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Re: There are 6 people at a party sitting at a round table with 6 seats: A  [#permalink]

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New post 24 Nov 2019, 12:40
VeritasKarishma

Hi

i m completely confused to infer the exact meaning of constraints of following questions.

1. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A cannot sit next to D and F at the same time. How many such
arrangements are possible?

2. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A can sit neither next to D nor next to F. How many such
arrangements are possible?

Kindly help me in getting over the confusion....

Per my understanding, the solution for both these questions are same.
But Per OA & your Combinatorics article, solutions are different as the constraints given are completely diff.

Please elaborate with an example.

Thanks in advance

Raju
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Re: There are 6 people at a party sitting at a round table with 6 seats: A  [#permalink]

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New post 24 Nov 2019, 23:46
1
gvvsnraju@12 wrote:
VeritasKarishma

Hi

i m completely confused to infer the exact meaning of constraints of following questions.

1. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A cannot sit next to D and F at the same time. How many such
arrangements are possible?

2. There are 6 people, A, B, C, D, E and F. They have to sit around a circular
table such that A can sit neither next to D nor next to F. How many such
arrangements are possible?

Kindly help me in getting over the confusion....

Per my understanding, the solution for both these questions are same.
But Per OA & your Combinatorics article, solutions are different as the constraints given are completely diff.

Please elaborate with an example.

Thanks in advance

Raju


Yes, they are different questions.

"A cannot sit next to D and F at the same time"
implies DAF is not possible and FAD is not possible. But AD (or DA) can sit together and FA (or AF) can sit together. Just that both F and D cannot sit next to A at the same time but one of them can sit next to A if the other does not.


"A can sit neither next to D nor next to F"
DA or AD is not possible. Also, AF or FA is not possible. Then of course, FAD or DAF is also not possible.
A cannot sit next to either of the two independently also.
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Re: There are 6 people at a party sitting at a round table with 6 seats: A   [#permalink] 24 Nov 2019, 23:46

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