How many six-digit numbers can be formed from distinct digits? Among these numbers, how many are there in which exactly four (no more, no less) odd digits appear?
A 6-digit number can be formed from 10 different digits, the number of which is the number of 6th class permutations without repetition, minus the number of groups starting with 0. The total number of permutations is $V_{10}^{6}$, one-tenth of which starts with 0, so in total

$$
\frac{9}{10} V_{10}^{6}=\frac{9}{10} \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=136080
$$

numbers can be formed that meet our first requirement.

From the digits $1,3,5,7,9$, we can choose a four-digit group in $\binom{5}{4}=5$ ways, without considering the order of the digits for now. Similarly, from the digits $0,2,4,6,8$, we can choose a two-digit group in $\binom{5}{2}=10$ ways, of which 4 groups contain the 0, and 6 groups do not contain 0. By combining each four-digit group with each two-digit group, we get $5 \cdot 4$ groups that contain a 0 and $5 \cdot 6$ groups that do not contain a 0. If we now permute the digits in each of these 6-digit groups, we get $5 \cdot 4 \cdot 6!$ groups that contain a 0, but only $\frac{5}{6}$ of these groups do not start with 0, and we get $5 \cdot 6 \cdot 6!$ 6-digit numbers that do not contain a 0. Therefore, the number of numbers that meet our second requirement is

$$
\frac{5}{6} \cdot 5 \cdot 4 \cdot 6! + 5 \cdot 6 \cdot 6! = 12000 + 21600 = 33600
$$