A and B get the same number of fish, so let's call that number $x$.
C has to get at least 2 fish, so let's call the number of fish C gets $y+2$.
Finally, D gets z fish, and there's no constraints on $z$.
So now, since the total number of fish is 11, we have $2x+(y+2)+z=11$, or $2x+y+z=9$. We're interested in the solutions where all three of $x$, $y$, and $z$ are nonnegative integers. Solving for $z$, $z=9-2x-y$, so any solution with $x$, $y$ nonnegative such that $2x+y \le 9$ will give a valid choice.
We can now just list these. For $x=0$, $y$ can be from $0$ to $9$, giving $10$ possibilities. For $x=1$, $y$ can be from $0$ to $7$. Similarly, for $x=2$, $y$ can go from $0$ to $5$, and for $x=3$, $y$ can go from $0$ to $3$. For $x=4$ the only possible values of $y$ are $0$ or $1$. So the number of ways is $10+8+6+4+2=30$.