thanx for draw.
fine, you first sit men and then you sit women between men
let's assume A B C D E F are different men and V X Y Z are women. When six men are seated there will be six slots available for women but women will occupy only four slots. For every ordered arrangement of six men you take 6P4 or simply said 360 ordered arrangements of women. Now my question is 'if A B C D E F are permuted, ordered 5! times (since circ. perm, due to clock and counter wise direction changes starting from the initial point) for every of 5! sets of men you assign 360 ordered arrangements of women? Your set A B C D E F is unique as one of total 5! arranged sets, and the assignment of woman set such as A-V B-X C-Y D-Z E F is the same as F E D-Z C-Y B-X A-V in circular order. If you had six women and six men then it wouldn't be like this, because all six possible slots between men were taken by women, but here it's repeating the same order of men-women. Therefore you over-count by six times each of the unique sets (ABCDEF), i.e. you assign six times more to each of 5! man sets.
Therefore, I first sit men like you do, then sit women in any of the four available slots (unlike you do) 5!*4! and for every combination of 5!*4! I'm taking additional 12 arrangements of two men reseated after six men and four women were seated. These two men are seated by each woman in different orders (they are permuted) A B C D E F -> A-V B-X C-Y D-Z E F -> A E F -V B-X C-Y D-Z -> A F E -V B-X C-Y D-Z E F. Please note when the order is turned backwards (counter-clock wise) I have not the identical orders, but for example two new ones -> F E A-V B-X C-Y D-Z and E F A-V B-X C-Y D-Z
5!*4!*4P2 against your order 5!*6P4, the difference is (5!*6) less than yours
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