Number of nine digits numbers whose sum of the digits is even I am reading Mathematical Circle. Problem $48$ in chapter two says that 

How many nine-digit numbers have an even sum of their digits?

I am trying in this way, that we can divide the problem in four cases. 


*

*$1$ even digit and $8$ odd digits

*$3$ even digits and $6$ odd digits

*$5$ even digits and $4$ odd digits

*$7$ even digits and $2$ odd digits


For the first case we get $4 \cdot 5^8 +5\cdot 5^7 \cdot 5$ number of solution. Because if the even digit is placed in first place (left to right) then we get $4\cdot 5^8$ ways to write the number and if an odd digit is placed in first place then we get $5\cdot 5^7 \cdot 5$ ways to write the number. Similarly for the second case we get $4\cdot 5^8+5^9$ , for the third case we get $4\cdot 5^8+5^9$ and for the fourth case we get $4\cdot 5^8+5^9$ ways to write the number. So total number is $ 4 \cdot (4\cdot 5^8+5^9)$. 
The answer is different. So Where I have made a mistake?
Thanks.

There are $900000000$ nine-digits numbers and exactly half of them have an even sum of digits (because every number can be paired with another of the opposite parity by changing the last digit).
