How many five-digit numbers can be formed $1^{\circ}$. from five identical digits, $2^{\circ}$. from two different, $3^{\circ}$. from three different, $4^{\circ}$. from four different, and $5^{\circ}$. from five different digits?
Below, we first calculate how many 1, 2, 3, 4, and 5-digit numbers can be formed from the digits 0 to 9. However, since numbers where the first digit is $0$ are not considered valid, the correct results are obtained by multiplying the results from the first row by $\frac{9}{10}$.

$1^{\circ}$. The number of 5-digit numbers that can be formed from 5 identical digits:

$$
\frac{9}{10} C_{1}(10)=9
$$

$2^{\circ}$. From two different digits, 5-digit numbers can be formed according to the following formulas: abbbb, aabbb, aaabb, aaaab. If $a$ and $b$ had only one value each (e.g., $a=1, b=2$), then the number of each group would be $\frac{5!}{4!}=5, \frac{5!}{2!3!}=10, \frac{5!}{3!2!}=10, \frac{5!}{4!}=5$, totaling 30. Since 10 digits allow for $\binom{10}{2}$ pairs to be chosen, the total number of groups is $45 \cdot 30=1350$. Considering that in one-tenth of the groups, 0 stands in the first position, the number of valid groups is: $\frac{9}{10} \cdot 1350=1215$. The result can be expressed in a formula as:

$$
\frac{9}{10} \cdot\binom{10}{2} \cdot\left[2 \cdot \frac{5!}{4!}+2 \cdot \frac{5!}{2!3!}\right]=1215
$$

$3^{\circ}$. Following the same procedure, we get:

$$
\frac{9}{10} \cdot\binom{10}{3} \cdot\left[3 \cdot \frac{5!}{3!}+3 \cdot \frac{5!}{2!2!}\right]=16200
$$

$4^{\circ}$.

$$
\frac{9}{10} \cdot\binom{10}{4} \cdot\left[4 \cdot \frac{5!}{2!}\right]=45360
$$

$5^{\circ}$.

$$
\frac{9}{10} \cdot\binom{10}{5} \cdot 5!=27216
$$

Adding up the results, we get 90000.

This number matches the total number of 5-digit numbers that can be formed from 10 digits; the fact that $\frac{9}{10} V_{5}^{i}(10)=90000$ also confirms the correctness of our method.

(László Bánó, Budapest.)

The problem was also solved by: Erdélyi J., Hajdu (Heimlich) P., Kirchknopf E., Kovács Gy.