+-649 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?
+32 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.
