You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.
Unless otherwise specified, in a circular arrangement, only the relative order matters.
Mum, Dad, and their six children ($3$ boys and $3$ girls) are to be seated at a circular table at random. What is the probability that all $3$ girls sit together?
Seat the mother first. Relative to her, the others can be seated in $7!$ ways as we proceed clockwise around the table.
For the favorable cases, we again seat the mother first. If the three girls sit together, we have five objects to arrange as we proceed clockwise around the table relative to the mother: the father, the three boys, and the block of three girls. The objects can be arranged in $5!$ ways. The three girls can be arranged within the block in $3!$ ways. Hence, the number of seating arrangements in which the three girls sit together is $5!3!$.
Thus, the probability that all three girls will sit together if the seats are randomly assigned is
$$\Pr(\text{all three girls sit together}) = \frac{5!3!}{7!}$$
Why was your solution incorrect?
As @gandalf61 pointed out in the comments, your answer
$$\frac{6!}{7!} = \frac{720}{5040} = \frac{5!3!}{7!}$$
is correct. However, your reasoning was incorrect.
You made two errors:
- You divided by $7! = 5040$ total arrangements, which suggests that you took relative order into account in the denominator. However, you obtained $6!$ arrangements of six objects in your numerator, which suggests that you did not take relative order into account in the numerator. If you take relative order into account in the denominator, you must take it into account in the numerator to be consistent.
- You also did not take into account the $3!$ ways the three girls can be arranged within the block.
Mum, Dad, and their six children ($3$ boys and $3$ girls) are to be seated at a circular table at random. What is the probability that Mum and Did sit opposite each other?
The denominator is the same as above.
For the favorable cases, seat the mother first. There is only one way to seat the father opposite the mother. Now seat the six children in the six remaining seats as you proceed clockwise around the table relative to the mother.
$$\Pr(\text{Mum and Dad sit opposite each other}) = \frac{6!}{7!}$$
Best Answer
Consider the $6$ girls as one person, then the total number of situations when $6$ girls sitting together is $7!$ multiplied by the permutations of girls $6!$, the total number of all permutations is $12!$, hence the probability is $\frac{6!7!}{12!}=\frac{1}{132}$.