How many 4-digit positive integers are there in which half of the digi

Total no. of ways in which out of 4 digits 2 are even and 2 are odd is \(\frac{4!}{2! * 2!}\) = 6 ways

The 6 cases are EEOO, EOEO, EOOE, OOEE, OEOE, OEEO. In these 3 cases start with Even numbers and 3 with odd numbers. Total no. of even numbers 0,2,4,6,8. Odd no.'s are 1,3,5,7,9.

1) EEOO = 4*5*5*5 = 500 numbers --> Since out of 5 even numbers 0 cant take first place so a total of 4 ways to fill that place and remaining each can be filled in 5 ways (as repetition is allowed)

2) EOEO = 4*5*5*5 = 500 numbers

3) EOOE = 4*5*5*5 = 500 numbers

4) OOEE = 5*5*5*5 = 625 numbers --> all places can be filled in 5 ways

5) OEOE = 5*5*5*5 = 625 numbers

6) OEEO = 5*5*5*5 = 625 numbers

Finally, 500*3 + 625*3 = 1500 + 1875 = 3375 numbers. I'm getting this answer. Is there any mistake in my approach? What is the correct answer and how?