9- 2014

The mistake that I made for my first try was that for some reason I neglected the zero condition.

Anyway straight forward counting:

first we place the zero, $3$ choices(not the first digit) then the odd number, $3$ choices for position and $5$ choices for number. finally the even number, $4*3$ ways to fill the vacant spots.

Together $3*3*5*4*3=\boxed {540}$ is our answer.

Let 4 digit number be ABCD

Case 1: A is odd

A = 5 possible odd numbers, B = 5 possible even numbers, C = 4 possible even numbers, D = 3 possible even numbers

Total = 5x5x4x3 = 300

Case 2: A is even

A = 4 possible even numbers, B = 5 possible odd numbers, C = 4 possible even numbers, D = 3 possible even numbers

Total = 4x5x4x3 = 240

Total Ways = 300 + 240 = 540

Call a, b, c and 0 are four digits of finding number. Having some formats of this number: abc0, ab0c, a0bc. In which a, b, c can be chosen from nine digits: 1, 2, 3, 4, 5, 6, 7, 8, 9. And the number of ways to choose abc from nine digits is: $C_3^{9}$ The number of ways to choose a odd digit is: $C_1^{5}$ The number of ways to choose two even digits: $C_2^{4}$ So the number of ways to get this number is: $C_1^{5} C_2^{4} \times 3 \times 3! = 5 \times \frac{4!}{2!^2} \times 3 \times 3! = \boxed{540}$