Count the number of ordered couples $(A, B)$ of subsets of $X$, with $|X| = 10$, such that $|A\cup B| = 3$ 
Count the number of ordered couples $(A, B)$ of subsets of $X$, with $|X| = 10$, such that $|A\cup B| = 3$

My thought was that there are 4 ways to have such two subsets:


*

*$|A| = 0, |B| =3$

*$|A| = 1, |B| =2$

*$|A| = 2, |B| =1$

*$|A| = 3, |B| =0$
For cases $(1), (4)$, we just have to choose $3$ elements of $X$, which can be done in $10\choose 3$ ways.
For cases $(2), (3)$, we have to choose $2$ elements for one subset and $1$ element for the other. The former can be done in $10\choose2$ ways, whereas for the latter there are $10$ possibilities.
Essentially, cases $(1),(2)$ mirror cases $(3),(4)$, therefore the answer is:
$$10{10\choose2}{10\choose3}+10{10\choose2}{10\choose3} = 20{10\choose2}{10\choose3}$$
Is my reasoning correct? I'm not totally sure about the part that involves the subset $\emptyset$. Do I have to count those cases as well?

You have not counted cases where $A\cap B$ is non-empty. To count all the cases correctly:


*

*Pick any three elements of $X$ to serve as $A\cup B$: $\binom{10}3$ ways

*For each of those three elements, decide whether it belongs in $A$ alone, $B$ alone or $A\cap B$: $3^3=27$ ways


In total there are thus $\binom{10}33^3=3240$ pairs.
