The sum of the six two-digit numbers is $(10a+b)+(10a+c)+(10b+a)+(10b+c)+(10c+a)+(10c+b)=22a+22b+22c$. Note that $\dfrac{484}{22}=22$, so $a+b+c=22$. Let us do casework on the largest digit among $a, b$ and $c$. Case 1: The largest digit is $9$. We have the following cases: (9,9,4) = 3 permutations (9,8,5) = 6 permutations (9,7,6) = 6 permutations for a total of $15$ permutations. Case 2: The largest digit is $8$. We have the following cases: (8,8,6) = 3 permutations (8,7,7) = 3 permutations for a total of $6$ permutations. Case 3: The largest digit is $7$ or less. The largest possible sum here is $21$, which is less than $22$, so there are no permutations for this case. This gives a grand total of $15 + 6 = \boxed{21}$ permutations in total.