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.
 A: 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).
A: This is an unnecessarily complex solution but shows some combinatorial techniques.
First, we'll count the number of (ordered) sequences of 9 digits where the sum of the digits is even. Then we will subtract the number of such sequences whose first digit is zero.
If the sum of the digits in the 9-digit sequence is even, then we must have an even number of odd digits (which can be 0, 2, 4, or 8). The number of such sequences is:
$$\sum_{n=0,2,4,6,8} {9 \choose n}\times{5^n}\times{5^{9-n}} =\sum_{n=0,2,4,6,8} {9 \choose n}\times{5^9}$$
Among these sequences, the number of the ones that start with $0$ making them non-9-digit numbers is:
$$\sum_{n=0,2,4,6,8} {8 \choose n}\times{5^8} $$
So the total number of 9-digit numbers whose sum of digits is even is:
$${5^8}\times\sum_{n=0,2,4,6,8} 5{9 \choose n}-{8 \choose n} = 450000000$$
A: $9$ digit numbers range from $10^8$ to $10^{9}-1$. There are exactly $9 \cdot 10^8$ numbers, that have nine digits.
$10^8 + 0$ doesn't have an even sum of its digits $(=1)$.
$10^8+1$ has an even sum of its digits $(=2)$.
$10^8+2$ doesn't have an even sum of its digits $(=3)$.
$10^8+3$ has an even sum of its digits $(=4)$.
$10^8+4$ doesn't have an even sum of its digits $(=5)$. 
$10^8+5$ has an even sum of its digits $(=6)$.
$10^8+6$ doesn't have an even sum of its digits $(=7)$. 
$10^8+7$ has an even sum of its digits $(=8)$.
$10^8+8$ doesn't have an even sum of its digits $(=9)$. 
$\vdots  \hspace{3cm}    \vdots \hspace{3cm}    \vdots$ 
$10^9-3$ has an even sum of its digits$(=-2)$.
$10^9-2$ doesn't have an even sum of its digits $(=-1)$. 
$10^9-1$ has an even sum of its digits $(=0)$. 
See the pattern? Here is a C++ Code that finds that out for you:
int main()
{
    unsigned long int x;
    x = 0;
    for(unsigned long int i = 100000000; i <= 1000000000-1; i++)
        if (i % 2 == 0)
            x++;
    std::cout << x;
}

This yields $x = (9 \cdot 10^8)/2 = 45\cdot 10^7$. Half of the numbers with nine digits have an even digit sum.
A: Let $A$ be the set of nine digit numbers. Its size is $|A|=9\cdot10^8$: we're free to choose eight digits among all ten digits, but the most significant digit can't be zero.
Now consider the following map $f\colon A\to A$:
$$
f(x)=\begin{cases}
x-9 & x\equiv 9\pmod{10} \\[4px]
x+1 & x\not\equiv9\pmod{10}
\end{cases}
$$
The map is bijective. If $A_0$ and $A_1$ denote the subsets of $A$ consisting of numbers with even or, respectively, odd digit sum, we can see that $f$ induces bijective maps $f_{01}\colon A_0\to A_1$ and $f_{10}\colon A_1\to A_0$.
Therefore $|A_0|=|A|/2=45\cdot10^7$.
