Probability: Unordered, with replacement In a state lottery four digits are drawn at random one at a time with replacement from 0 to 9. Suppose that you win if any permutation of your selected integers is drawn. Give the probability of winning if you select: 6,7,8,9.
I am most confused about the different ways to approach this problem. In thinking about it with "order does matter," there are 4! possible orderings of 6, 7, 8, and 9. Furthermore, there are 10^4 ways to get all possible outcomes. That is the correct answer, 4!/(10^4). 
My professor said that order doesn't matter. So the possible wins are 1 because there's only one possible set of 6,7,8,9. Next, the possible outcomes are (10^4)/4! - all the outcomes divided by the number of ways to arrange those outcomes, to get unordered outcomes. That, too, is numerically correct, because it works out to 4!/(10^4).
But if order really doesn't matter, and you can have replacement, why would you not use n+r-1 C r for the possible outcomes? In this approach, there is one possible set, so it would be 1/(10 C 4), but that deviates from the right answer. I don't understand where my problem is.
 A: The number of possible outcomes is not $10^4/4!$, that's not even an integer. In order to count outcomes, we need to first define what we mean by outcome, that is, we have to describe the sample space.  
Also, even though ultimately order does not matter, the number of outcomes is not $\binom{10}{4}$. For since drawing was done with replacement, it is possible to get repetition of digits. And if we did give a real count of possible outcomes, that included for example the possibility of getting three $1$'s and one $4$, we would still have some work to do. For the event three $1$'s and one $4$, and the event one each of $3,6,8,9$ are not equally likely.  
It is convenient (though not always possible) to set up a set of possible outcomes such that all outcomes are equally likely. Then we can find our probability by counting. 
If we do that, then if our counting is correct, we will get the right answer. In this particular problem, the sample space of $10^4$ strings of length $4$ consists of equally likely outcomes. Since we are counting order, for the numerator we have to use $4!$.
