I'm struggling on a question similar to the following:
Your ID can only use 3 digits. Allowed digits are 0-8. How many allowed combinations are there if repeating digits is allowed? What about if they're not allowed?
I feel like it has something to do with nPr and nCr but I really am not sure. Any help would be very much appreciated.
On
In the first case, the first digit can be anything from $1$ to $9$. The same applies for the other two digits. Hence there are $10$ digits to choose form in each place. So the total number of cases will be $9{\times}9{\times}9=729$
Now when digits cannot be repeated, the first digit can be anything from $1$ to $9$ ($9$ options). But the second digit can be anything except the first digit which gives use $8$ options. Similarly, we have $7$ options for the third digit. Total number of cases $9{\times}8{\times}7=504$
When repeating digits are not allowed this is a permutation because order matters. You have to choose 3 without repeating and the order matters.
$_9P_3 = \frac{9!}{(9-3)!} = 504$
When repeating digits is allowed you have 9 choices for each digit and it doesn't matter what the previous digits were so you can just use the multiplication principle.
$9^3 = 729$