Six people sit at a circular table. Three men and three women. What is the probability that all girls sit together?
I start by identifying the cardinality of the sample set (omega). It's a circular table so rather than having the normal $6!$ permutations we divide by $6$ it to $5!$. We have three men and three women. Their gender is their only identifying factor so they're interchangeable. This is true for both so we now have $\frac{5!}{3!3!}$.
Here's the problem. When this resolves I get $\frac{10}{3} \approx 3.333$ repeating. What am I doing wrong to end up with a fraction?
Assuming that had resolved as an integer I'd move to find the cardinality of E, our Event. For our event all three woman act as one unit leaving 3 interchangeable men and one group of woman, or 4 objects. It's a circular table so we divide by $4$ and get $3!$. We still have three Thus we could find the cardinality of E by simplifying $\frac{3!}{3!}=1$.
They are not interchangeable: if the women are $W_1,W_2$, and $W_3$, a seating arrangement in which they sit in the order $W_1W_2W_3$ is distinguishable from one in which they sit in the order $W_2W_1W_3$.
Since the table is circular, we can list any arrangement by starting with $W_1$ and then going around the table clockwise from her. As you say, there are $5!$ ways to fill in the other $5$ people. In order to get the $3$ women seated together, we must have one of the arrangment patterns $W_1WWMMM$, $W_1WMMMW$, or $W_1MMMWW$, where $W$ stands for a woman and $M$ for a man. Thus, there are $3$ pairs of seats in which we can put $W_2$ and $W_3$, and we can seat them in either order. Finally, there are $3!$ possible seating orders for the $3$ men, so there are altogether $3\cdot2\cdot3!=36$ arrangements that have all $3$ women sitting together, and the probability of getting one of them is $\frac{36}{5!}=\frac3{10}$.
Or we can be a little cleverer and realize that if the $3$ women are together, the $3$ men must also be together. The $3$ women can be seated in $3!$ different orders in $3$ consecutive seats, and the $3$ men can then fill the remaining $3$ seats in $3!$ different orders. This yields the same total of $3!\cdot3!=36$ acceptable arrangements and the same probability of $\frac3{10}$ of getting one of them.