What is the amount of different four-digit numbers that can be created from the digits $0, 1, 2, 3, 4$ and $5$? 
What is the amount of different four-digit numbers that can be created from the digits $0, 1, 2, 3, 4$ and $5$?

The solution is $431,$ but I have no idea how this solution was found. How can I solve this problem?

As stated in some of the comments, the problem is in the description of the question. We can distinguish the following cases:


*

*Using the digits to form a four-digit number. The first digit cannot equal $0,$ while all digits can be used for the second, third and fourth digit. As such, the number of possibilities equals: $$5 \cdot 6^3 = 1080$$

*Using the digits to form a number which contains four unique digits. The first digit cannot equal $0,$ while the second digit cannot equal the first, the third digit cannot equal the first or the second and the fourth digit cannot equal the first, the second or the third. As such, the number of possibilities equals: $$5 \cdot 5 \cdot 4 \cdot 3 = 300$$

*Using the digits to form a number which contains at most four unique digits. We can distinguish four cases for the number of digits, and the number of possibilities thus equals: $$5 \cdot 5 \cdot 4 \cdot 3 + 5 \cdot 5 \cdot 4 + 5 \cdot 5 + 6 = 431$$
The latter scenario is the one considered by the authors, although the original question was badly posed. It should in fact have read:

What is the amount of numbers that can be created from the digits $0, 1, 2, 3, 4$ and $5,$ if the number cannot contain more than four digits and every digit can be used at most once?

