Use the number 0123456 to form a four digit number without repetition, in which the number of one digit, ten digit and hundred digit only have a few with the even four digit number

Use the number 0123456 to form a four digit number without repetition, in which the number of one digit, ten digit and hundred digit only have a few with the even four digit number

C61*A63-A33-C31*C42*A33=606
Starting from the thousand digits, you can't use 0, so you can choose one of 123456, so it's C61; then you can choose three of the remaining six digits in order, so it's a63; you can get the answer by subtracting one digit, ten digit, and the number on the hundred digit,
==Sorry, it seems that the above is wrong ~! Let's write it again
C31*(A43+C41*A33)+C31*（A33+C32*C31*A33）=324
It should be the answer ~!
I have two situations: one is that C31 starts with odd numbers 1, 3, and 5, and then C31 is followed by even numbers of one digit, ten digit, and hundred digit. There are two small situations: (1) the last three digits are all even (including 0), which are 0, 2, 4, and 6, so A43; (2) the last three digits are one even and two odd, and one even (including 0), which is C41. Two odd numbers need not be selected, but only 3 and 5, and then they are arranged, That is, multiply by A33. Synthesis: C31 * (A43 + C41 * A33)
The other is to start with even numbers 2, 4 and 6. Similarly, C31 can be divided into two small cases: (1) the last three digits are all even (including 0), so there are only three left, and A33 is directly arranged; (2) the last three digits are one even and two odd, one even (including 0) is C31, two odd is C32, and then A33 is arranged
Total synthesis: C31 * (A43 + C41 * A33) + C31 * (A33 + C32 * C31 * A33) = 324
PS: if I'm right, it should be 324