[TwoaDay]

Four couples enter a restaurant. How many ways can they be seated at a round table so that the men and women alternate and no husband and wife sit next to each other?

[Guest]

ok, that was random, was expecting a site or something...lmao

how about scanning and posting the page out of your book, i don't feel like spending $$$ to see the answer given

don't know if this causes problems with copyright or anything though, oh well, thanks for the laugh

[TwoaDay]

lol we must have been typing at the same time

[TwoaDay]

are you guys happy now, i'd like to see you try to prove me wrong now

[Guest]

I can't help but feel like I'm being trolled here. The picture from the book is below.

You can't see any other way to do this?

[Guest]

  Tearz said:

Four couples enter a restaurant. How many ways can they be seated at a round table so that the men and women alternate and no husband and wife sit next to each other?

It doesn't ask for seat placement. If they did, then I would have to agree with you.

Lets take option number 1 and 21, the 8 people are seated exactly the same way, just in different seats.

Are you saying that if you went to a table with 8 seats (by yourself) and sat down, there is only one possible way you can sit?? And because there is only one "way" every person whom sits at the table will ALWAYS sit in the same seat? I dont think so!

[Guest]

these are 12 distinct sitting arrangements all with man one starting in the same location then you could do the same 12 for each of the other 7 chairs allowing 96 seperate sitting arrangements in all

table12.bmp

[TwoaDay]

its the order, not the seats, of course the can sit in different seats, ther are just only 2 orders

[Guest]

  finance_it said:

these are 12 distinct sitting arrangements all with man one starting in the same location then you could do the same 12 for each of the other 7 chairs allowing 96 seperate sitting arrangements in all

I agree with you!

[TwoaDay]

agree with him all you want your both wrong

we need someone really smart to clear this up

[Guest]

  TwoaDay said:

agree with him all you want your both wrong

we need someone really smart to clear this up

Here is 12 unique solutions

1 M1 W2 M3 W1 M4 W3 M2 W4 3056

2 M1 W2 M3 W4 M2 W1 M4 W3 3098

3 M1 W2 M4 W1 M3 W4 M2 W3 3178

4 M1 W2 M4 W3 M2 W1 M3 W4 3194

5 M1 W3 M2 W1 M4 W2 M3 W4 3656

6 M1 W3 M2 W4 M3 W1 M4 W2 3698

7 M1 W3 M4 W1 M2 W4 M3 W2 3892

8 M1 W3 M4 W2 M3 W1 M2 W4 3920

9 M1 W4 M2 W1 M3 W2 M4 W3 4370

10 M1 W4 M2 W3 M4 W1 M3 W2 4424

11 M1 W4 M3 W1 M2 W3 M4 W2 4492

12 M1 W4 M3 W2 M4 W1 M2 W3 4520

prove to me that 10 do not meet the outlined conditions

[Guest]

better graphic showing the 12 seating positions, notice couples never sit next to each other and it is not just a twist of the seating positions there are 12 distinctly different positions, and as before solid circle never moves to another seat, so there would be 96 total positions available if solid circle would move to each of the 8 starting positions with the same setup as in these 12, i believe your book is flawed

[TwoaDay]

whatever i dont care any more, i still think theres two but dont really know or care

agree to somewhat disagree

[Guest]

Great diagrams.

I do agree with the 96 and 12 answers.

If the puzzle asked:

"Four couples enter a restaurant, how many ways can they be seated at a round table so that the men and women alternate"

For this question I believe you can have two answers:

1. Using a standard linear permutations approach where your outcome is 96, where each seat is distinguaishable from the others.

2. Using a circular permutations approch where your outcome is 12, where the seats are indistinguishable, so long as each person has the same two people each side.

However, because the puzzle asks"

"Four couples enter a restaurant. How many ways can they be seated at a round table so that the men and women alternate and no husband and wife sit next to each other?"

I believe the answer is 2.

Why?

if we take husband 1 and wife 1 there are only two positions around the table they can sit from each other, across and to one side (the side where they would not be sitting next to each other). Given these two seating options (and using circular permutation) you could say that the other 3 couples could then arrange themselves a few ways around couple 1. But they can't because they each have only two positions to sit in relation to their partner around the table as well.

And this is where I believe the answer is 2, because of the added condition where husbands and wives cannot sit next to each other.

Lets take again husband 1 and wife 1, we seat them first and now there are three ways the other 3 couples can seat themselves around couple one.

Now everyone stands up and we take couple 2 and seat them, there are three ways the other 3 couples can seat themselves around couple 2. The same for couples 3 and 4. So we take 4 couples and multiply it by 3, this gives us 12. However, we don't ground one couple at a time and rearrange the other 3 around them, they all have to postition themselves around the table together at the same time and in doing so there are only two ways they can all do this together.

Edited July 30, 2008 by Tearz

[Guest]

Oh and sorry for my spelling errors, I'm sure you brainiacs will know what I'm sayng

[TwoaDay]

thank you tearz

[Guest]

  Tearz said:

Great diagrams.

I do agree with the 96 and 12 answers.

If the puzzle asked:

"Four couples enter a restaurant, how many ways can they be seated at a round table so that the men and women alternate"

For this question I believe you can have two answers:

1. Using a standard linear permutations approach where your outcome is 96, where each seat is distinguaishable from the others.

2. Using a circular permutations approch where your outcome is 12, where the seats are indistinguishable, so long as each person has the same two people each side.

However, because the puzzle asks"

"Four couples enter a restaurant. How many ways can they be seated at a round table so that the men and women alternate and no husband and wife sit next to each other?"

I believe the answer is 2.

Why?

if we take husband 1 and wife 1 there are only two positions around the table they can sit from each other, across and to one side (the side where they would not be sitting next to each other). Given these two seating options (and using circular permutation) you could say that the other 3 couples could then arrange themselves a few ways around couple 1. But they can't because they each have only two positions to sit in relation to their partner around the table as well.

And this is where I believe the answer is 2, because of the added condition where husbands and wives cannot sit next to each other.

Lets take again husband 1 and wife 1, we seat them first and now there are three ways the other 3 couples can seat themselves around couple one.

Now everyone stands up and we take couple 2 and seat them, there are three ways the other 3 couples can seat themselves around couple 2. The same for couples 3 and 4. So we take 4 couples and multiply it by 3, this gives us 12. However, we don't ground one couple at a time and rearrange the other 3 around them, they all have to postition themselves around the table together at the same time and in doing so there are only two ways they can all do this together.

Thanks for the explination, I see what you are trying to do.

The combinations which finance_it adn Myself have provided, there are no husband and wife sitting next to each other, so it meets the same criteria as the puzzle.

However, I think, where we differ is in our interperation of the puzzle. You interperate it as looking at an indivual couple, where as I am looking at a seating plan senerio.

[TwoaDay]

well ive decided that the answer is 2 or 12 depending or your interpretation

( i still thinks its two)

[Guest]

  TwoaDay said:

well ive decided that the answer is 2 or 12 depending or your interpretation

Im willing to agree with you there.. ( i still think 96)

[Guest]

Well I am willing to agree with everyone. It is how you interpret the question.

So thats a yes to 96, 12 and 2

[Guest]

12 :-)

that diagram makes most sense to me