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

The (moderately) less elegant approach that the OP (i.e. original poster) attempted is also feasible.
If the leftmost digit is odd, then the number of satisfying digits is
$$5 \times \binom{4}{1} \times 5^4. \tag1 $$
If the leftmost digit is even, then the number of satisfying digits is
$$4 \times \binom{4}{2} \times 5^4. \tag2 $$
Adding the expressions in (1) and (2) above gives
$$5^4 \times \left[ ~(5 \times 4) + (4 \times 6) ~\right] = 27500.$$
