How many four-digit positive integers are there such that all digits are different?
For a valid $4$-digit number, the Highest Significant digit will be among $1\cdots9$ so can have $9$ values
From the next onwards, they can assume $0\cdots 9$ excluding the ones already selected
So, the number of combinations will be $$9\cdot9\cdot8\cdot7$$
The first digit can be any number from $1$ to $9$. The next three digits can be any number from $0$ to $9$. If we choose each digit one at a time and stipulate that each digit is different from the previous choices, how many choices do we have for each digit? Multiply these together to get your answer.
Allowing for a zero first digit
10 * 9 * 8 * 7
10 possible options for the first digit, 9 remaining options for the second digit, 8 remaining options for the third digit ...