How many permutations of [9] have no adjacent odd digits?

First look at the same problem for [5] we have:

12345 12543 14325 14523

32145 32541 34125 34521

52341 52143 54321 54123

This suggests that for each odd digit we have a permutation of the rest of the even digits times the permutation of the rest of the odd digits (I think) giving

$\displaystyle

3*\left(2\right)_{2}*\left(2\right)_{2}

$ = 12

Then the answer to the original question would be

$\displaystyle

9*\left(4\right)_{4}*\left(4\right)_{4}

$ = 5184

Does this solution hold any merit or am I totally confused as usual?

Thanks