How many five-digit numbers can be formed

a) entirely from equal digits,

b) from two different,

c) from three different,

d) from four different,

e) from five different digits?
In each case, we determine the number of groups that can be formed from 10 digits according to the conditions. $\frac{9}{10}$ of this number will give us the number of $s z a ́ m o k$ (numbers) we are looking for, because we need to exclude the groups starting with 0 (which constitute $\frac{1}{10}$ of all groups).
a) From 10 digits, we can choose 1 digit in $C_{10}^{1}$ ways. From the chosen digit, only 1 group of 5 elements can be formed. Therefore, the number of 5-digit numbers consisting of the same digit is

$$
\frac{9}{10} C_{10}^{1}=\frac{9}{10}\binom{10}{1}=\frac{9}{10} \cdot 10=9
$$

b) From 10 digits, we can choose 2 different digits in $C_{10}^{2}$ ways. Denoting the chosen 2 elements by $a, b$, the groups of 5 elements that can be formed are the permutations of $a b b b b, a a b b b, a a a b b, a a a a b$. The number of these permutations is $2\left(P_{5}^{4}+P_{5}^{3,2}\right)$. Therefore, the number of 5-digit numbers containing 2 different digits is:

$$
\frac{9}{10} \cdot\binom{10}{2} \cdot 2\left(\frac{5!}{4!}+\frac{5!}{3!2!}\right)=\frac{9}{10} \cdot \frac{10 \cdot 9}{1 \cdot 2} \cdot 2(5+10)=81 \cdot 15=1215
$$

c) For 3 different digits, the following groups can be formed: $aaabc, a b b b c, a b c c c, a a b b c, a b b c c$. The number of these groups is $3\left(P_{5}^{3}-1-P_{5}^{2,2}\right)$, and thus the number of all 5-digit numbers formed from different digits is:

$$
\frac{9}{10} \cdot\binom{10}{3} \cdot 3\left(\frac{5!}{3!}+\frac{5!}{2!2!}\right)=\frac{9}{10} \cdot \frac{10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3} \cdot 3(20+30)=324 \cdot 50=16200
$$

d) Similarly, the number of 5-digit numbers that can be formed from 4 different digits is:

$$
\frac{9}{10} \cdot C_{10}^{4} \cdot 4 \cdot P_{5}^{2}=\frac{9}{10}\binom{10}{4} 4 \cdot \frac{5!}{2!}=\frac{9}{10} \cdot \frac{10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4} \cdot 4 \cdot 60=756 \cdot 60=45360
$$

e) From 5 different digits,

$$
\frac{9}{10} C_{10}^{5} P_{5}=\frac{9}{10}\binom{10}{5} 5!=9 \cdot 9 \cdot 8 \cdot 7 \cdot 6=27216
$$

5-digit numbers can be formed. (This number is also $\frac{9}{10}$ of the number of 5th class variations that can be formed from 10 elements:

$$
\frac{9}{10} V_{10}^{5}=\frac{9}{10} 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6=27216
$$

Note: The total number of 5-digit numbers in the above a), b), c), d), e) is the total number of existing 5-digit numbers, which is $90000\left(99999-9999\right.$ or $\left.\frac{9}{10} V_{10}^{1,5}=\frac{9}{10} \cdot 10^{5}=90000\right)$, and indeed $9+1215+16200+45360+27216=$ 90000.

Number of incorrect solutions: 14. Apart from not considering all 10 digits, the most common mistake was working with repeated variations, forgetting that among the $k$-class repeated variations formed from $n$ elements (even if $k>n$) there are groups that do not contain all $n$ elements, and in some cases only 1 element. Furthermore, they did not exclude the groups starting with 0.