# round table

• February 18th 2013, 05:24 AM
kastamonu
round table
4 married couples sit at a round table. How many ways can they sit if each husband must sit opposite his wife?

• February 18th 2013, 06:18 AM
Plato
Re: round table
Quote:

Originally Posted by kastamonu
4 married couples sit at a round table. How many ways can they sit if each husband must sit opposite his wife?

Seat couple A at the table.
Now there are six ways to seat couple B.
Then there are four ways to seat couple C.
Finally there are two ways to seat couple D.
• February 18th 2013, 06:57 AM
kastamonu
Re: round table
3!.16 . If we think four couples as one person we can place them in 3! ways. 2 couples will be ordered between themselves and there are 4 couples this makes 16. But your nswer is better. Many Thanks.
• February 18th 2013, 07:17 AM
Plato
Re: round table
Quote:

Originally Posted by kastamonu
3!.16 . If we think four couples as one person we can place them in 3! ways. 2 couples will be ordered between themselves and there are 4 couples this makes 16.

I don't know where you are getting your model, but it is wrong.
Draw a regular octagon. There vertices are the eight chairs.
It is not ordered until one couple is seated.
Seat couple A anywhere, but man and wife are at opposite vertices.
Now the table is ordered.

There are six seats left. Seat wife B in any one of them. Then man B has only one place to be seated, at the opposite vertex from wife B.

$6\cdot 4\cdot 2=48$.
• February 18th 2013, 09:17 AM
kastamonu
Re: round table
What happens if we want all the couples to sit seperately?
• February 18th 2013, 09:24 AM
kastamonu
Re: round table
There are 4 married couples. If we think them as 1 person they can be arranged in 3! They will change places between themselves in 2! ways. TRhere are 4 couples and this makes 2!2!2!2!3!
• February 18th 2013, 11:00 AM
Plato
Re: round table
Quote:

Originally Posted by kastamonu
What happens if we want all the couples to sit seperately?

Always, always, start a new thread for a new question.

Using inclusion/exclusion $\sum\limits_{k = 0}^4 {\left( { - 1} \right)^k \binom{ 4}{k} \cdot 2^k \left( {7 - k} \right)!}$