2 boys and 4 girls stand in a circle. Count the ways in which 2 boys and 4 girls can stand in a circle. (rotations of the same pattern are identical). 
I am manually counting 3 patterns for this. I do not know if there is a general formula that we can come up with. 
$\textbf{[Harder]}$ Count the ways in which 10 different boys can stand in a circle. (rotations of the same pattern are identical). 
 A: This is only a partial solution, but too long for a comment.  Suppose there are $b>1$ boys and $g>1$ girls.  Fix some position that will have a boy in it.  The remaining children can be arranged in the circle, counterclockwise from the starting boy, in $$\frac{(g+b-1)!}{(b-1)!g!}$$ ways.  However, we can rotate any of the boys to the staring position, and have the same cyclic arrangement, so each pattern has been counted $b$ times.  This gives $$\frac{(g+b-1)!}{b!g!}$$ patterns.
The problem with this is that some rotation may leave the arrangement fixed, so that there is no double counting.  If we try the above formula with $g=4, b=2$ we get $$\frac{5!}{2!4!}=\frac52.$$  The problem is the the arrangement BGGBGG, when rotated gives exactly the same arrangement.  The correct answer is given by $$\frac12\left(\frac{5!}{1!4!}-1\right)+1=3$$ 
To solve the problem in general, we have to work out how many rotations there are that leave the arrangement fixed.  In general, it may be that some of the rotations leave the pattern fixed, and some don't.  In the pattern,
BGBGGGBGBGGG
rotating by $5$ leaves the pattern fixed, but rotating by $2$ does not.
I doubt there's a general formula.  you probably have to investigate each case separately.  I imagine that Polya's theory of counting comes into it.    
A: Assuming that for the first question the boys are indistinguishable, and same for the girls, then yes, there are $3$ patterns: there can be $0$, $1$, or $2$ girls between the two boys.
You may want to ask your instructor though if indeed the boys are all regarded as the same, and same for the girls ...
For the second question: Here are two ways to think about that one:
First: What are the possible number of line-ups of $10$ different boys?  So when you make the line into a circle, and get rotational symmetry, then how many line-ups would end up becoming the same circle?  Divide the number of lines-ups by that number and you have the number of different circles.
A second method is this: take one of the $10$ boys, and place this boy anywhere at the table. By rotational symmetry, it does not matter where this first boy ends up.  However, once this first boy is seated, it kind of 'anchors' the table. That is, depending on who gets seated to the right of the first boy, you get a different arrangement. In fact, there are $9$ boys left who can be seated to the right of the first boy.  And now there are $8$ left who can be seats to the right of the second boy, etc. So from this you can get the right formula as well.
