# Seating arrangement

• September 30th 2010, 07:16 AM
Mondreus
Seating arrangement
Four married couples are having dinner together. The seating is arranged in such a way that the four ladies draw lots to see which man they end up sitting with.

What is the probability that exactly two wives get to sit with their own husband? The answer is 1/4, but I would like to see a solution involving permutations and/or combinations if possible.
• September 30th 2010, 07:37 AM
Plato
Quote:

Originally Posted by Mondreus
Four married couples are having dinner together. The seating is arranged in such a way that the four ladies draw lots to see which man they end up sitting with.
What is the probability that exactly two wives get to sit with their own husband? The answer is 1/4, but I would like to see a solution involving permutations and/or combinations if possible.

There are $4!=24$ possible ways to seat the women.
There are $\dbinom{4}{2}=6$ ways to seat exactly two with their partner.
• September 30th 2010, 08:47 AM
Mondreus
Shouldn't the husbands be selected in order as well? I understand the basics of permuations and combinations, but I'm having a hard time figuring out how to piece it together.
• September 30th 2010, 08:56 AM
Plato
Quote:

Originally Posted by Mondreus
Shouldn't the husbands be selected in order as well? I understand the basics of permuations and combinations, but I'm having a hard time figuring out how to piece it together.

No, there is no reason to consider the husbands other than each identifies a position. Think of four two-person tables. Seat the men in any order at each table. Now you have four ‘named’ positions into which we put the women.
• September 30th 2010, 10:42 AM
Mondreus
The answer says $\frac{\text{number of ways to fixate two couples} \times \text{the remaining possibilities}}{\text{number of ways to place the husbands}}$
• September 30th 2010, 10:49 AM
Plato
Quote:

Originally Posted by Mondreus
The answer says $\frac{\text{number of ways to fixate two couples} \times \text{the remaining possibilities}}{\text{number of ways to place the husbands}}$

How does that give $\frac{1}{4}$ as an answer?

I totally disagree with that reasoning.