The number of arrangements that can be formed with the letters of the word ORDINATE, so that the vowels occupy odd places is ( Count the places from right to left)

The set of letters of the word ORDINATE is $\left\{O,R,D,I,N,A,T,E\right\}$

The vowels are $A,E,I,O$ and the consonants are $R,D,N,T$

There are 8 places , the odd places are $1,3,5,7$ these four places can be filled with four vowels in $4!$ ways, the remaining four places can be filled w$4!$ith consonats, in $4!$ ways

Therefore, the number of of arrangements that can be formed with the letters of the word ORDINATE, so that the vowels occupy odd places is $4!·4!=\boxed{576}$