"$4$ boys and $3$ girls sit round a table. What is the probability that exactly $2$ girls will sit next to each other?"
I said the following: there are ${7\choose3} = 35$ ways to pick places for the girls, and in $5$ of those ways, the girls will be completely apart (i.e. no two girls sit together), giving a probability of $\frac{5}{35} = \frac{1}{7}$. Thus the probability that at least two girls sit together is $1-\frac{1}{7} = \frac{6}{7}$. The probability that all three girls sit together is $\frac{7}{35} = \frac{1}{5}$, since there are $7$ ways to pick the three places such that all three girls sit together. Hence, the probability that exactly two girls sit together is $\frac{6}{7} - \frac{1}{5} = \frac{23}{35}$.
However, I'm told that the correct answer should be $\frac{3}{5}$, so my question is: what is wrong with my working above?
If the seats are $(A,B,C,D,E,F,G)\,$ then we can have them all separate using $$(A,C,E),\,(A,C,F),\, (A,D,F),\,(B,D,F),\,(B,D,G),\,(B,E,G), \,(C,E,G)$$
Thus there are $7$ ways, not $5$ as stated in the post.
I think, however, that it is easier to do the problem directly. There are $7$ places to start the pair of girls (working clockwise, say). Then, having placed the consecutive girls, there are $3$ places to put the third (as we exclude the occupied seats as well as the two adjoining seats). Thus there are $7\times 3=21$ ways to place them such that exactly two are consecutive. It follows that the answer is $$\boxed {\frac {21}{35}=\frac 35}$$