# Circular table permutations problem

• May 11th 2013, 06:57 AM
PJani
Circular table permutations problem
Hey, i have problem understanding how to get result for circular permutations where i have repeating elements.
I have 10 people in circle, 3 man and 7 women. How can i get number of permutations?

Result in book is 36.
I tried
(9!/2!+7!)+(9!/3!+6!)=120 which is wrong.
• May 11th 2013, 07:16 AM
Plato
Re: Circular table permutations problem
Quote:

Originally Posted by PJani
Hey, i have problem understanding how to get result for circular permutations where i have repeating elements. I have 10 people in circle, 3 man and 7 women. How can i get number of permutations?

Result in book is 36.

That problem as you have posted it is absolutely incomplete.

As stated the answer is simply $9!$.

There must be many more conditions placed on the groups of men and women.

Please add in those further conditions. That is the only way the answer $36$ makes any sense.
• May 11th 2013, 07:26 AM
PJani
Re: Circular table permutations problem
Heh sory, my mistake!
How can i get number of permutations where i distinguish persons only by gender?
• May 11th 2013, 07:50 AM
Plato
Re: Circular table permutations problem
Quote:

Originally Posted by PJani
Heh sory, my mistake!
How can i get number of permutations where i distinguish persons only by gender?

In the future, please try to be clear and complete.

Place a female anywhere at the table.
Now you have a string $ffmmmmmmm$ to arrange: $\frac{9!}{2!\cdot 7!}=?$ ways.

Start at the seated female's right and seat each of those strings counter-clockwise.
• May 11th 2013, 08:01 AM
PJani
Re: Circular table permutations problem
Why do you start with females? And why is $\frac{9!}{2!\cdot 7!}=?$ and not $\frac{9!}{3!\cdot 6!}=?$