Probability that $1,2$ are in the same set on randomly halving $\{1,...,10\}$? 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.
 A: 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 $\frac49$.
A: 
The total number of partitions would be 10-choose-5. If 1 and 2 are in the same set, then then there are 8-choose-3 ways left to create the set that contains them since two of the numbers have already been chosen.

$\binom{10}5$ counts the ways to select a set of five elements from ten. But do notice that each such selection will also create an additional set of five elements: its relative complement.
There are $\tbinom 22\tbinom 83$ ways to select elements 1, 2, and three from the remaining eight elements into a set of five elements.  However, there are also $\tbinom20\tbinom 85$ ways to select five from those eight elements which will leave 1 and 2 in the complement set.  The favoured event is the union of both these disjoint cases.
$\def\lfrac#1#2{\left.{#1}\middle/{#2}\right.}$So the probability we seek is $\lfrac{\left[{\tbinom22\tbinom 83+\tbinom20\tbinom 85}\right]}{\tbinom {10}5}$ which equals $\lfrac{2\tbinom 83}{\tbinom{10}5}$ or simply $\lfrac 4 9$.
We can validate this by noting that there are ${\tbinom21\tbinom84}$ ways to select one from $\{1,2\}$ and four from the eight other elements, counting all ways to place 1 and 2 in separate sets, so the probability we seek is $\lfrac{\left[\tbinom{10}5-\tbinom21\tbinom84\right]}{\tbinom{10}5}$, which is indeed $\lfrac 49$.
Unsurprisingly, since $\tbinom{10}5=\tbinom 20\tbinom85+\tbinom 21\tbinom84+\tbinom 22\tbinom83$ .

Alternatively, as @BrianTung notes, there are $\tbinom 94$ distinct ways to partition the ten element set into two sets of five elements.  This counts ways to select one set that contains a particular element and one which does not.  We can choose any of the ten elements to be the set identifier, but it is quite useful to use 1.
The count of partitions where 2 is in the set containing 1 (the favoured event) is $\tbinom 8 3$. So the probability we seek is $\lfrac{\tbinom 8 3}{\tbinom 9 4}$, which is $\lfrac 49$.
(Note: We get the same answer if we choose 2 as the identifying element.)

We might make things harder for ourself and choose another element as the identifier; such as 10.  In this case we should note that the favoured event is satisfied when 1 and 2 are both in the identified set, or when they are both in its complement.
So the probability we seek is $\lfrac{\left[{\tbinom22\tbinom 72+\tbinom20\tbinom 74}\right]}{\tbinom 94}$, which again equals $\lfrac 49$.

A different approach might be to count distinct ways two special positions might placed when the ten elements are lined up, which is $\tbinom{10}2$.  The favoured event is when those two positions are both among the first five, or the last five.  Those positions are for elements 1 and 2.
So the probability we seek is $\lfrac{\left[\tbinom 52\tbinom 50+\tbinom 50\tbinom 52\right]}{\tbinom{10}2}$.  This is unsurprisingly equal to $\lfrac 49$.

However you decide to count equally probable atoms of the total space, then the atoms of the favoured event should be counted in the same general manner.  Also make sure you add the counts for all disjoint cases that comprise the favoured event.
Remember too that sometimes counting the disfavoured atoms might be easier.
