, 2019

Solution:

$A=4$, $B=7$. There are $28$ total ways of choosing an element from $U$ and $V$, so there must be $14$ ways where $U$'s is larger. If we relabel the elements to be $0,1, \cdots, 10$, then element $i$ is greater than exactly $i$ elements in the set. However, we overcount other elements in $U$, so the four elements in $U=\{a, b, c, d\}$ must satisfy
$$(a-0)+(b-1)+(c-2)+(d-3)=14 \Rightarrow a+b+c+d=20$$
To remove the uniqueness condition, we subtract $1$ from $b$, $2$ from $c$, and $3$ from $d$, so we wish to find solutions $a \leq b \leq c \leq d \leq 7$ to $a+b+c+d=14$. From here, we do casework. If $a=0$, $b=0,1,2,3,4$ give $1,1,2,2,3$ solutions, respectively. If $a=1$, $b=1,2,3,4$ give $2,2,3,1$ solutions, respectively. If $a=2$, $b=2,3,4$ give $3,2,1$ solutions, respectively. If $a=3$, the only solution is $3,3,4,4$. Thus, the answer is $(1+1+2+2+3)+(2+2+3+1)+(3+2+1)+1=24$.