Twelve balls numbered $1$ through $12$ are placed in a bin. Joe produces a list of three numbers by performing the following sequence three times: he chooses a ball, records the number, and does not replace the ball in the bin. How many different lists are possible?
If order doesn't matter, then there are C(12, 3) choices, which is \((12 \cdot 11 \cdot 10) \div (3 \cdot 2 \cdot 1)=1320 \div 6=220.\)
However, if order does matter, we count the amount of choices for the first option:12. The second option has 11 choices, and the third, 10. we multiply them to get 1320 choices.