How many five-digit numbers are there that have exactly two odd digits in their decimal notation? I thought that in the first place you can choose 9 out of 10 numbers (all but zero) then in the second place choose 4 out of 5.
Or the second option in the first place can be 4 out of 5 odd numbers, in the second and third places we choose odd.
it seems to me that I am confused in these combinations, I don’t understand how to calculate, given that zero cannot be the first

As suggested in the comments, take the $\binom{5}{2}5^25^3$ numbers with at most five digits, and subtract the numbers with at most four digits, that is $\binom{5}{2}5^5-\binom{4}{2}5^4=44\cdot 5^4=27500$.
