ANSWER.  (4*5)*(5*4*3*2) = 20*120 = 2400.


Explanation


(1)  We can use any one of 4 even digits and can place it to any of 5 possible positions.

     It gives the factor 4*5 = 20.



(2)  Next we should place remaining 5 odd digits (1, 3, 5, 7 and 9) in remaining 4 positions. 

        (Notice that after placing the even digit in #(1), its position and remaining positions are just 
        determined by a unique way - they are not a subject of choice anymore).
    

     Therefore, placing in #(2) can be done by 5*4*3*2 = 120 ways.



(3)  In total, it gives  20*120 = 2400 different ways, in accordance with the answer above.