Count the possible ways to seat people at a round table I have a very simple combinatorics problem, and I am (almost) sure I did this right. However, the smartest guy in my class thinks different and this made me doubt, so I will post our views to this problem and I would like to know who is right.

In how many different ways can 10 people be seated around a round table, under the condition that there are 5 man and 5 women, and no two males or two females can be seated adjacently.

My view
We number the chairs in an arbitrary way, so the chair numbering is fixed. At chair $1$, there is seated either a man or a woman, which given $2$ possibilities. When we observe the gender of the person in the first chair, the division of the table over the two genders in given. Then there are $5!$ ways to seat the men over the $5$ male chairs and $5!$ ways to seat the women over the $5$ female chairs. So in total this gives $2.(5!)^2$ possibilities.
My classmates view
First all males are seated, which is possible in $5!$ ways. Because no males can sit adjacently and the people are seated in a circle, we can rotate the table and in this way we count all possibilities 5 times too much, so $5!/5=4!$ ways to seat the man. After this, the table division is fixed, but we still have to seat the women. This gives another $5!$ ways, so in total $5!4!$ ways.
What I think is wrong with this view
I don't think the argument about rotation of the table makes any sense. The argument can make sense though, if we use exactly my classmate's argument, but acknowledge that the first person to be seated can be any of the $10$ people, so giving an extra $10$ possibilities. Then proceed as my classmate, giving the same answer as me. However, my classmate does not agree. So who is right?
Also, if we use our arguments to the same problem but a table of $2$ people, one male, one female, there are obviously $2$ possibilities. Then my method gives $2.(1!)^2=2$ possibilities, but my classmate's method gives only $1!0!=1$ possibility.
EDIT 
I think the main problem is in different interpretations of the problem, where my interpretation is that it matters who sit at 'chair 1' while my classmates thinks that only the ordering of the people matters. However, given only the above problem, what would be the right interpretation, or is this ambiguous?
 A: Your disagreement comes from different interpretations of the question. 
You're assuming the chairs around the table are indexed ${1,...,10}$ and you're looking at the number of ways to assign the numbers ${1,...,10}$ to your 5 men and 5 women.
Your classmate is looking at how many ways you can arrange 5 men and women around the table relative to each other instead of in absolute terms.
The difference between the two answers is a factor of 10 ($10\times4!5! = 2\times 5 \times 4!5! = 2\times(5!)^2$), which corresponds to the extra degree of freedom (e.g. where is chair #1).
A: The thing is, that when you are a circle, each combination $x_1, x_2 \dots x_{10}$ is counted $10$ times, namely 
$x_1, x_2 \dots  x_9 x_{10}$, 
$x_2, x_3 \dots x_{10},x_1$, 
$x_3, x_4 \dots x_1, x_2$,
$\dots$
$x_{10}, x_1, \dots x_8, x_9$
since rotating the table preserves the combination of people sitting
In the case of $2$ people you get the combinations:
man, woman
woman, man
but they are essentially the same.
Your edit is spot on, depends on whether the chairs differ from one another, but it'd assume that this is not the case
A: In combinatorics, the definition of "different" is often a crucial issue. There is nothing in the problem saying that the chairs are interchangeable, and you should not add assumptions to the problem. If I were only one person to be seated, I would have ten choices where to sit. Your interpretation is the correct one. If your classmate's interpretation were intended, then the problem is poorly stated. 
And actually, if we to follow your classmate's reasoning, then why stop there? Your classmate is noting that rotation leaves the arrangement "essentially" the same, but what about reflections? If all that matters is who is sitting next to whom, and absolute location is unimportant, then wouldn't the answer be 5!4!/2?
A: Your solution would be okay if there is a "special chair" (you name it "chair $1$"). Then there is no essential difference with placing the $10$ persons in a row in which case there is also a special chair (for instance the utmost left, or the utmost right). 
If there is no special chair then you can start by placing one man. After that there are $4!$ different arrangements for the other men and $5!$ different arrangements for the women. This leads to a total of $4!5!$ possibilities.
It is not for nothing that a round table is used, indicating that there is no special chair. So I would say that your classmate is right. 
