How many six-digit numbers can be formed with six different digits? Among these numbers, how many have four odd digits?
$1^{\circ}$. As many different six-digit numbers can be formed as the number of variations without repetition of six elements from ten, that is,

$$
v_{6}(10)
$$

However, since we cannot consider the numbers whose first digit is 0, we must subtract one-tenth of the number of groups, so the number of the sought six-digit numbers is

$$
v_{6}(10)-\frac{1}{10} v_{6}(10)=\frac{9}{10} \cdot\binom{10}{6} \cdot 6!=136080
$$

$2^{\circ}$. If a six-digit number contains four odd digits, the other two digits must be even. The four combinations that can be formed without repetition from the digits $1,3,5,7,9$ are: $1357,1359,1379,1579,3579$, and the two-combinations that can be formed without repetition from the digits $0,2,4,6,8$ are: $02,04,06,08,24,26,28,46,48,68$. If we now connect each of the two-combinations with each of the four-combinations, we get $5 \cdot 4=20$ six-combinations where 0 occurs and $5 \cdot 6=30$ six-combinations where 0 does not occur. If we also consider that the elements in the thus obtained six-combinations can change places and that we cannot consider the groups where 0 is in the first place, then we get that the total number of groups is

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