Permutations with Repetitions, how to select $n$ and $r$ I'm doing Permutations and Combinations and I'm facing difficulty with Permutations with repetitions. My book states that if there are $n$ objects chosen $r$ at a time, the total number of permutations is $n^r$.
I cannot understand how to decide what $n$ is and what $r$
In the following problem:

In how many ways can 6 different rings be worn on 4 fingers?

While I can solve this using the Multiplication Principle, I keep thinking that $n$ is the number of rings (as in my textbook it says that there are '$n$' different things') and $r$ is the number of fingers.
 A: It is reasonable to suppose that the order of rings on fingers matters. Blue then white then red on the ring finger looks better than red then blue then white.
To count the number of ways to choose the number of rings the various fingers will hold, use Stars and Bars. The number of choices is $\binom{9}{3}$.  For each choice, the rings can be permuted in $6!$ ways, for a total of $\binom{9}{3}6!$.
A: Definition.  A permutation with repetition (or $r$-tuple or word) is an ordered selection of $r$ elements from a set of $n$ elements in which repetition is allowed.
By the Multiplication Principle, the number of words of length $r$ that can be formed with a set of $n$ elements is $n^r$ since we have $n$ choices for each of the $r$ times we make a choice. 
Example. Consider the number of sequences of five letters we can form from the English alphabet if repetition is permitted.  Since there are $26$ letters in the English alphabet, we have $26$ choices for each letter.  This is our $n$.  The number of times we choose a letter is $5$, which is our $r$.  Thus, there are $26^5$ sequences of five letters of the English alphabet if repetition of letters is permitted. 
Observe that $n$ is the number of choices, while $r$ represents the number of times you make that choice.
Andre Nicholas has pointed out in the comments and his solution that the number of ways that six different rings can be placed on four fingers is not a permutation with repetition (as I mistakenly thought) since the rings are not identical, so the order in which they are placed on the fingers matters.    
A: I would recommend that you think concretely instead of worrying about what is $n$ and $r$. Effectively that would mean that you derive the formula $n^r$ each time you solve the problem but in the long run that would be helpful.
You say that you can solve this using the multiplication principle, but the formula $n^r$ is only an abstract (special case) of the multiplication principle, so you are implicitly using the formula anyway. With experience the abstraction part would come naturally and forcing it is not good IMHO.
In this case just think about the rings one by one. The first ring has four choices (fingers) to go to, the second has four as well and so on, and that would give you the answer $4^6=4\times 4\times...\mbox{six times}$.
