Count the possible ways to seat people at a round table

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.

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).

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