Textbook problem: The integers from $1$ to $10$ inclusive are partitioned into two sets of five elements each. What is the probability that $1$ and $2$ are in the same set?
My solution: $2/9$. The total number of partitions would be $10 \choose 5$. If $1$ and $2$ are in the same set, then there are $8 \choose 3$ ways left to create the set that contains them since two of the numbers have already been chosen. Hence, the probability is ${8 \choose 3} / {10 \choose 5}$ which is $2/9$.
Textbook answer: $4/9$.
Question: What am I missing in my reasoning? Should I be doubling the number of ways to create a set with $1$ and $2$ in it for some reason?
Textbook: The Art Of Problem Solving (Vol. 1) by Rusczyk, Chapter 26.
Best Answer
One way of looking at your error is that, yes, there are $\binom{10}{5}$ five-element subsets altogether, but only half of them contain $1$. Of those that contain $1$, $\binom{8}{3}$ also contain $2$. So the desired probability is
$$ \frac{\binom{8}{3}}{\frac12\binom{10}{5}} = \frac{56}{\frac{252}{2}} = \frac{4}{9} $$
Alternatively, you can observe that whichever subset $1$ is in contains four of the remaining nine numbers, so the probability that $2$ is one of those four is $\frac49$.