How many integers from 1-999 do not have any repeated digits?
The answer is explained in this link, but why is the last set 9*9*8? Why not 9*9*9?
How many integers from 1-999 do not have any repeated digits?
The answer is explained in this link, but why is the last set 9*9*8? Why not 9*9*9?
In the link, they claim that $9\cdot 9\cdot 8$ gives the number of integers from $100$ to $999$ with no repeated digits. You have three digits $d_1d_2d_3$ to work with. Begin with the first digit $d_1$: it can be any number in $\{1, \ldots, 9\}$. When choosing the second digit $d_2$ you must avoid repeating $d_1$, but you can now use $0$, so you select from $\{0, \ldots, 9\} - \{d_1\}$. So while the first two factors of $9 \cdot 9 \cdot 8$ are the same, there is an interesting difference between picking $d_1$ and picking $d_2$. Similarly, the the last digit must come from $\{0, \ldots, 9\} - \{d_1, d_2\}$, and this set has size $8$.