How many $5$-digit zip codes are there with exactly $n$ odd digits?

If zip codes have $5$ digits and each digit can be $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$, then how many zip codes are there with exactly $n$ odd digits?

The way I see it, there are $10^5$ possibilities total.

For exactly one odd digit, we need one odd digit and four even digits. Which makes the answer $5^5$ but this can't be right, because according to this same logic $5^5$ is the answer for all other $n$'s too!

EDIT: And this also doesn't take position into account. I.e. 52222 is not the same as 25222 and etc.

• Firstly, there's no digit '10' in any zipcodes, but there are still $10^5$ possibilities. First, find all arrangements of 4 even numbers. Then insert 1 odd number. – Ian Coley Apr 8 '14 at 22:26
• One odd: The location of the odd can be chosen in $\binom{5}{1}$ ways. For each of these ways, the oddball can be chosen in $5$ ways. And now the rest of the slots can be filled with evens in $5^4$ ways. Two odd: Locations can be chosen in $\binom{5}{2}$ ways. The two locations can be filled with odds in $5^2$ ways. And the rest can be filled with evens in $5^3$ ways. – André Nicolas Apr 8 '14 at 22:32

First we look at the zipcodes using only $o$'s and $e$'s, where $o\in \{ 0,2,4,6,8\}$ and $e\in \{1,3,5,7,9\}$.
There are $5 \choose n$ ways to make a zipcode for which the number of $o$'s is $n$.
So for instance, there are $5$ zipcodes with one $o$ (e.g. $eeeeo$).
Then we say: For each $e$ and each $o$ there are five possibilities. This means that there are a total of ${5\choose n} 5^5$ ways to make a $5$-digit zip code with $n$ odd digits.
Check: $$\sum_{k=0}^5{5\choose k}5^5=5^5\sum_{k=0}^5{5\choose k}=5^5(1+5+10+10+5+1)=5^5\cdot 32=5^5\cdot 2^5=10^5 \tag{\checkmark}$$