3B. We will say that a number is "fancy" if it is written with an equal number of even and odd digits. Determine the number of all four-digit "fancy" numbers written with different digits?
Solution. We have 5 even digits: $0,2,4,6,8$ and 5 odd digits: $1,3,5,7,9$. Two even digits out of 5 can be chosen in $C_{5}^{2}=10$ ways, i.e., we have the following possibilities

$$
\{\{0,2\},\{0,4\},\{0,6\},\{0,8\},\{2,4\},\{2,6\},\{2,8\},\{4,6\},\{4,8\},\{6,8\}\} .
$$

In the same number of ways, two out of 5 odd digits can be chosen.

For each pair (one pair of even and one pair of odd numbers), there are a total of $4!=24$ ways to form a sequence of four digits. Therefore, we have a total of $10 \cdot 10 \cdot 24=2400$ four-digit sequences consisting of two even and two odd digits.

It remains to count those that start with 0. If a pair of even numbers contains 0, then the number of four-digit numbers starting with 0 is $3!=6$. Since we have four even pairs containing 0, it follows that the number of all "special" four-digit sequences starting with 0 is $4 \cdot 10 \cdot 6=240$.

Therefore, according to this, the number of all four-digit "special" numbers written with different digits is $2400-240=2160$.