How many five-digit numbers can be formed from the digits $1,2,3, \ldots, 9$ that contain three odd and two even digits, if a) we allow the repetition of digits, b) we do not allow the repetition of digits.
I. Let's first consider the case where digits cannot be repeated. Imagine writing down all the odd and even digits of each suitable five-digit number separately, keeping both parts in the order found in the number; for example, 72196 would yield the components $7,1,9$ and 2,6. We will determine how many ways the odd component of our numbers can be formed, how many ways the even component can be formed, and how many ways these two components can be combined into a five-digit number.

The first digit of the odd component can be chosen in 5 ways from 1, 3, 5, 7, and 9, the second digit in 4 ways from the remaining 4 digits, and the third digit in 3 ways, so the number of different odd components is $5 \cdot 4 \cdot 3=60$. Similarly, the number of even components is $4 \cdot 3=12$.

For the combination, it is sufficient to specify which 2 of the 5 positions will be filled with the digits of the even component, for example, in the above example, the 2nd and 5th positions, with the remaining positions filled with the digits of the odd component. The first selected position can be chosen in 5 ways, the second in 4 ways, and the resulting $5 \cdot 4=20$ pairings are identical in pairs, for example, selecting the 5th and then the 2nd position results in the same pair of positions. Thus, the number of different pairs of positions is $20: 2=10$.

By recombining each odd component with each even component in all 10 position distributions, we always get a different five-digit number, and thus we get all the desired numbers. Therefore, excluding repetitions, the number of suitable numbers is $60 \cdot 12 \cdot 10=7200$.

II. Similarly, it follows that allowing repetition, $(5 \cdot 5 \cdot 5) \cdot(4 \cdot 4) \cdot 10=20000$ suitable numbers can be formed.