# one more question

• September 23rd 2008, 10:00 AM
chrisc
one more question
I am not sure how to begin this one. Total outcomes would be 11!, so i was assuming with Mr and Mrs A sitting opposite 10!, but I think that is just them sitting apart. You will see what I mean.

Two couples (Mr&Mrs B
and Mr&Mrs B)and 8 other people are to be
randomly seated at a round table. In how many of the possible
11! circular
arrangements will Mr and Mrs A
sit directly opposite each other and (at the

same time - this is
one question) Mr and Mrs Bwon’t sit next to each other.

Thanks
• September 23rd 2008, 10:59 AM
Soroban
Hello, Chris!

I think I've solved it . . .

Quote:

Two couples, Mr & Mrs A and Mr & Mrs B and 8 other people are to be randomly seated
at a round table. In how many of the possible 11! circular arrangements will Mr and Mrs A
sit directly opposite each other and Mr and Mrs B do not sit next to each other.

Mr. A can take any seat . . . it doesn't matter.
Mrs. A must take the opposite seat (one choice).
The other 10 people can be seated in $10!$ ways.
. . Hence, there are 10! ways that the A's are directly opposite each other.

Of these 10! arrangements, how many have the B's together?

The A's are already seated opposite each other.

Duct-tape the B's together.
Then we have 9 "people" to seat.
There are 9! ways for them to be seated.
. . Hence, there are 9! ways for the B's to be together (and the A's are opposite).

Therefore, there are: . $10! - 9! \:=\:3,265,920$ ways
. . that the A's are opposite and the B's are not together.

• September 23rd 2008, 01:30 PM
Plato
Quote:

Originally Posted by Soroban
Duct-tape the B's together.
Then we have 9 "people" to seat.
There are 9! ways for them to be seated.
. . Hence, there are 9! ways for the B's to be together (and the A's are opposite).

Should that not be $2(9!)$?
Husband B can sit to the right hand of husband A only if husband B is on the left hand of wife B. So you must double that count.
• September 23rd 2008, 02:02 PM
bubbie
Hey guys,

What I am not sure about is if there are twelve people altogether (2+2+8=12), how can we have 11! possible arrangements in total? Would not it be 12!? Am I missing something?

Now, if take the couple (Mr. and Mrs. A) as being fixed – that is, occupying two PARTICULAR chairs, placed directly opposite of each other – we have ten people and ten chairs left, which implies that these remaining ten people can be seated in 10! different ways. Moreover, the couple (Mr. and Mrs. A) can be seated in 12 different ways, yet remain directly opposite of each other, right? Therefore, we can conclude that Mr. and Mrs. A can be seated directly opposite of each other in 12*10! different ways. Do you follow my drift? Or am I completely confused? :-)

If we now move on to Mr. and Mrs. B not sitting next to each other, we follow Soroban’s logic of duct-taping them together. This, indeed, implies that now the “nine” people can be seated in 9! various ways. However, we are forgetting one minor detail here: these calculations assume that Mr. and Mrs. B are sitting in the exact same way, for instance Mr. is always seated to the right of Mrs. Thus, we have to take into account that they can be also switched and that would double the number of possibilities – i.e. 2*9!

Hence, following this thought, the total number of possible arrangements meeting the imposed conditions is 12*10! – 2*9! = 2*9!*(6*10-1)=118*9!
• September 23rd 2008, 02:06 PM
bubbie
Quote:

Originally Posted by bubbie
Hey guys,

What I am not sure about is if there are twelve people altogether (2+2+8=12), how can we have 11! possible arrangements in total? Would not it be 12!? Am I missing something?

Now, if take the couple (Mr. and Mrs. A) as being fixed – that is, occupying two PARTICULAR chairs, placed directly opposite of each other – we have ten people and ten chairs left, which implies that these remaining ten people can be seated in 10! different ways. Moreover, the couple (Mr. and Mrs. A) can be seated in 12 different ways, yet remain directly opposite of each other, right? Therefore, we can conclude that Mr. and Mrs. A can be seated directly opposite of each other in 12*10! different ways. Do you follow my drift? Or am I completely confused? :-)

If we now move on to Mr. and Mrs. B not sitting next to each other, we follow Soroban’s logic of duct-taping them together. This, indeed, implies that now the “nine” people can be seated in 9! various ways. However, we are forgetting one minor detail here: these calculations assume that Mr. and Mrs. B are sitting in the exact same way, for instance Mr. is always seated to the right of Mrs. Thus, we have to take into account that they can be also switched and that would double the number of possibilities – i.e. 2*9!

Hence, following this thought, the total number of possible arrangements meeting the imposed conditions is 12*10! – 2*9! = 2*9!*(6*10-1)=118*9!

This looks like too many possibilities though (Nerd)
• September 23rd 2008, 02:13 PM
icemanfan
The number of possible arrangements will be dependent on whether a certain arrangement of people simply rotated (everyone sits a certain number of chairs to the right of their original position) is considered to be the same arrangement or a different arrangement.
• September 23rd 2008, 02:21 PM
Plato
Quote:

Originally Posted by bubbie
What I am not sure about is if there are twelve people altogether (2+2+8=12), how can we have 11! possible arrangements in total? Would not it be 12!? Am I missing something?

You are missing the fundamental property of circular arrangements.
If we arrange n distinct objects is a circle as opposed to arrangements in a line (queue) we must divide by n,
$\frac {n!} {n} = (n-1)! \;$, to account for the n identical rotations.
• September 23rd 2008, 02:40 PM
bubbie
Quote:

Originally Posted by Plato
You are missing the fundamental property of circular arrangements.
If we arrange n distinct objects is a circle as opposed to arrangements in a line (queue) we must divide by n,
$\frac {n!} {n} = (n-1)! \;$, to account for the n identical rotations.

Point taken :-)
• September 23rd 2008, 04:42 PM
bubbie
Had a couple more minutes to spend on the question and here is my final thought :-)

If the chairs are indistinguishable, the A couple, including ten other people, can be seated in $10!$ different ways. The point here is that the chairs are, in fact, not numbered – that is, the seating arrangement is relative to people and not chairs.

Furthermore, the B couple can be seated together in $2(9!)$ various ways.
Therefore, the "right" solution to the initial problem should be $10! - 2(9!) = 8(9!) = 2,903,040$
• September 30th 2008, 02:21 PM
bubbie
Well, the right answer was $10!-8(2)(8!)=2,983,680$