Step 1: Translate each condition into an equation.
Call the five picked integers, sorted from smallest to largest, $a \le b \le c \le d \le e$, each between $0$ and $20$.
Mean $12$ means $a+b+c+d+e = 60$. Median $18$ means the middle value $c = 18$. A single mode of $20$ means $20$ appears more times than any other number in the list, with no other number tying it.
Step 2: Work out where the value 20 can sit.
Because the list is sorted and $c = 18$, we know $a \le b \le 18$ and $18 \le d \le e \le 20$. The value $20$ is bigger than $18$, so it can only appear in the $d$ or $e$ position, never in $a$, $b$, or $c$.
For $20$ to be a mode at all it must repeat, appearing at least twice; the only two "slots" that could ever hold $20$ are $d$ and $e$, so both must equal $20$: $d = e = 20$.
Step 3: Reduce to a two-variable problem.
Plug $c = 18$, $d = 20$, $e = 20$ into the sum:
$a + b + 18 + 20 + 20 = 60 \Rightarrow a + b = 2$, with $0 \le a \le b \le 18$.
Listing all integer solutions with $a \le b$: $(a,b) = (0,2)$ or $(1,1)$.
Step 4: Test the single-mode condition on each candidate list.
For $(1,1)$: the list is $1,1,18,20,20$. Counting repeats, $1$ appears twice and $20$ appears twice, a tie, so there is no single mode; this fails the question's requirement.
For $(0,2)$: the list is $0,2,18,20,20$. Counting repeats, $0$, $2$, and $18$ each appear once, while $20$ appears twice, strictly more than any other value, so $20$ is the unique mode. This list passes every requirement.
Step 5: Count the valid lists.
Only the multiset $\{0,2,18,20,20\}$ satisfies mean $=12$, median $=18$, and a single mode of $20$ at the same time. Since we ignore permutations (different orderings of the same five numbers do not count as different picks), this is exactly one valid way to choose the five integers.
Step 6: Final Answer.
\[ \boxed{1} \]