$3$ Boys and $3$ Girls sit at a circular table. What is the probability the boys sit together? Three boys and three girls sit at a circular table. No boy or girl is more likely to sit on a particular chair. What is the probability the three boys all sit together?
I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving $5!$ possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy ($\times2$) and the next boy can sit to the right or left of him ($\times2$).
And as the $3$ boys can be arranged in $3!$ ways this gives $2\times2\times3!$ possibilities for the boys out of $5!$ overall. However this doesn't seem to work as the answer is $3/10$. This would fit with $\frac{3\times2\times3!}{5!}$ but I cant see why to change one of these twos into the required three. Am I missing something?
Many thanks for any help - I am sure it is something obvious I am overlooking so feel free to say so
 A: Call the first boy A, second boy B, and third boy C. Your method doesn't allow for the ordering ACB.
The correct logic is as follows: make the left-most boy sit in a fixed chair - so we have BBB GGG. Then there are a total of $(3!)^2$ ways of making such an arrangement. Divide this by $5!$ and you get $3 \over 10$.
A: 
I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving 5! possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy (×2) and the next boy can sit to the right or left of him (×2).

The first problem, is that this only counts the number of solutions where the first two boys are sitting next to each other.
The second problem, is that you are going between (a) counting how many ways to place the boys in the numerator and (b) counting how many ways to place the everyone, boys and girls, in the denominator.  Either is fine, but you have to pick one and stick with it.
So if we only count how to place boys: there's $4 \times 5$ ways total to place the boys, without them needing to be together.  With all 3 boys together:  there's $4$ ways to place them with first and second next to each other and only $2$ ways to place them so that first and second are not next to each other.  So $(4+2)/(5 \times 4) = 3/10$.
If we count how to place everyone, that's $5!$ ways in general.  There's $4$ ways to place the first two boys next to each other like before, but then $3!$ ways to place the remaining girls.  There's $2$ ways to place the first two boys not next to each other, like before, but still $3!$ ways to place the remaining girls.  So $(4\times 3! + 2\times 3!)/5! = 3/10$.
A: The boys can be placed in ${6\choose3}$ ways. There are exactly $6$ seatings where the boys sit together. Your probability therefore is
$${6\over{6\choose3}}={3\over10}\ .$$
