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
I have tried to find a 'neat' solution. But cannot. So this is messy.
Originally Posted by qwerty10
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.
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"