We have four even digits  0, 4, 8, 8  and 4 odd digits  3, 3, 7, 7.


Two possible placements are  EOEOEOEO  or  OEOEOEOE,  where E is the placeholder 
for an even digit and O is the placeholder for an odd digit.


For odd digits with repetitions, as in the given number, there are

     =  = 6 different distinguishable permutations are possible 

inside the group of odd numbers.


For even digits with repetitions, as in the given number, there are

     =  = 12 different distinguishable permutations are possible 

inside the group of even numbers.


Of these 12 arrangements, the number of those that start with 0 is 3;  so these 3 should be subtracted from 12.
Doing this way, we find that the number of arrangements of four even numbers in this problems, 
that do not start from 0 is 12-3 = 9.



Thus we have 6*12 = 72 arrangements of the type OEOEOEOE and 9*6 = 54 arrangements of the type EOEOEOEO.


The total is the sum 72 + 54 = 126.



Hence, in total, there are  126 different distinguishable arrangements as required in the problem.   ANSWER