Combinatorics, are the people indistinguishable or not? I'll try my best to translate so if something doesn't seem right, please do tell and I'll try my best to rephrase.
How many ways can you sit 2 kids, 6 women and 6 men down around a circular table so that kids sit together and men don't sit next to men and women don't sit next to women.
Firstly, I think that the question asks just about formations of people and the people are not "unique" (so basically it doesn't matter if John sits on the first chair or if he switches seats with Mark because it is the same formation).
Secondly, because it is a circular table you can move the table around so the formation is always the same. (Something I had mentioned, If everyone is seated and then everyone changes one seat to the left it is still the same way because the table is circular [or am I crazy?])
Therefore, if I am correct in this, the answer should be two (because one formation is with a woman next to the kid and the second is with a man next to the kid).
If the people are unique and it matters whether John and Mark sit on the first and third seat or vice-versa, then I think it should be 2*6!*6!*2 (how I got this is two kids can be seated in a way kid1 - kid2 or kid2 - kid1 and then one of six men next to one of six women next to one of five men next to one of five women etc. times two because it next to the kid can be one of six women next to one of six men next to one of five women next to one of five men)
Lastly, If it doesn't matter that the table is circular (it can't be moved around) and the people are unique it should be the previous times 14 because the kids can be seated in 14 different seat combinations (12, 23, 34, 45, 56, 67, 78, 89, 910, 1011, 1112, 1213, 1314, 141)
Thank you for any feedback :)
 A: As a rule, unless specifically mentioned otherwise, for seating at a round table

*

*people are taken as distinguishable

*with unnumbered seats, arrangements remain unchanged under rotation

Thus, for this particular problem,

*

*seat the two kids in two ways (clockwise) $AB$ or $BA$ as reference group.

*seat women-men /men-women alternating clockwise from the reference group

*both the women group and men group can be permuted in their gender specific seats in $6!$ ways

Putting it all together,
number of arrangements = $2\times 2\times  6!6!$
A: Unless otherwise specified, people are considered to be distinguishable.
By convention, in a circular arrangement, only the relative order of the people matters unless the seats are labeled.
Suppose the children are Anne and Brian.  Seat Anne.  We will use her as a reference point.  There are two ways to seat Brian next to Anne.  Once they are seated, fill the remaining seats by proceeding clockwise around the table from the block of children.  Choose whether a man or a woman sits immediately to the left of the block of children.  This determines which seats will be occupied by men and which seats will be occupied by women.  Once that choice is made, there are $6!$ ways to arrange the six men in the seats reserved for men and $6!$ ways to arrange the six women in the seats reserved for women.  Hence, there are
$$2 \cdot 2 \cdot 6!6!$$
distinguishable seating arrangements of two kids, six men, and six women if the two children sit next to each other, no two men are adjacent, and no two women are adjacent.
If the seats were labeled, you are correct that we would have to multiply the above answer by $14$ since there would be $14$ distinguishable ways to seat Anne.
