circular permutations

• March 26th 2011, 10:08 AM
qwerty10
circular permutations
3 girls and 3 boys can be seated at a circular table so that any two and only two of the ladies sit together. How many ways can they be seated?

My method was: 2 girls have to be seated toegther they can be arranged in 2! ways. You then have 1 girl and 3 boys left, 2 of the boys have to sit either side of the first 2 girls we placed as there cannot be 3 girls together. Those 2 boys can be places in 2! ways either side of the girls. There are then 2 places left where those places can be ordered in 2! ways. Giving 2x2x2=8 ways as the answer- which is way off the correct answer of 72, yet I cannot find an alternative way of thinking about it
• March 26th 2011, 11:26 AM
Plato
Quote:

Originally Posted by qwerty10
3 girls and 3 boys can be seated at a circular table so that any two and only two of the ladies sit together. How many ways can they be seated?

I have tried to find a 'neat' solution. But cannot. So this is messy.
There are three ways to select the lady to be alone.
The are three ways to seat a man to her right and two ways to seat a man to her left.
There are two ways the seat the other two ladies and two place to seat them.
Thus $(3)(3)(2)(2)(2)=72$.
• March 26th 2011, 11:41 AM
qwerty10
Thank you for the help
In selecting the first girl who sits on her own, aren't we in effect selecting someone as head of the table, so taking that person as given? Then we wouldnt have 3 possibilities as that person wouldnt count in the sum. For example, when we have to order 6 people around a cirular we can select one person as "head of the table" then there are 5! possibilites of ordering the other people around this "head of table"
• March 26th 2011, 12:39 PM
Plato
Quote:

Originally Posted by qwerty10
In selecting the first girl who sits on her own, aren't we in effect selecting someone as head of the table, so taking that person as given? Then we wouldnt have 3 possibilities as that person wouldnt count in the sum. For example, when we have to order 6 people around a cirular we can select one person as "head of the table" then there are 5! possibilites of ordering the other people around this "head of table"

No that is in fact wrong. She does count because she is alone. So there are three ways to pick her.
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Here is a second way to work it.
There are $(6)(6)=36$ ways to seat all the men and all the women together.

There are two ways to seat the women at the table in every other chair. Then there are six ways to seat the men in the remaining chairs.
That is 12 ways to seat this party so not two people of the same sex are next to each other.
That is there are $36+12=48$ ways to have all sexes together or apart. So there are $5!-48=72$ ways to have exactly two ladies together.