In how many ways 5 different rings can be worn on 4 fingers? Is the answer $4^5$ or $4 \times 5 \times 6 \times 7 \times 8$?
The rationale behind $4 \times 5 \times 6 \times 7 \times 8$ is that first ring has four options, the second ring has $5$ options, since the second ring can come underneath the first, and so on.
I think $4^5$ is correct since $4 \times 5 \times 6 \times 7 \times 8$ is over counting the number of possibilities.
 A: Five identical rings can be placed on four fingers (using Stars and Bars) in $\binom{5+4-1}{4-1}=\binom{8}{3}$ ways.
Because the rings are not identical, they can then be permuted, in $5!$ ways.
So the answer is:
$$\binom{8}{3}5!=\frac{8!}{3!}=4\cdot5\cdot6\cdot7\cdot8$$.
Which happens to be $P(8,5)$, interpreted as:

Pick 5 letters from ABCDEFGH. The remainder are the bars, the selection is mapped alphabetically to 12345 for the rings.

A: Apparently, each finger needn't have a ring, and in this context, $4\cdot5\cdot6\cdot7\cdot8$ is indeed correct.
An easy to understand way from the diagrams below where $F$ represents finger, $R$ ring and taking that a ring is attached to a finger to its right, initially we have
$-F-F-F-F$ and the first ring has $4$ places to get inserted
On inserting a ring in one of the gaps, we have. say,  $-F-F-R-F-F$
which clearly shows that there are now $5$ gaps for the next ring
Also try placing the next ring beside a placed ring, $-F-F-R-R-F-F$
and see that there are now $6$ gaps for the next ring
Each time you place a ring anywhere, an additional gap gets created,
thus the answer is $4\cdot5\cdot6\cdot7\cdot8$
A: You solution is very clever. I believe $4 \times 5 \times 6 \times 7 \times 8$ is indeed correct. The logic is sound.
To answer your question, there is no double counting. If there were double counting, there would be two paths that yielded the same outcome. But if each ring is sequentially placed in the correct relative order as your original method suggests, any alternate path would create two rings in an incorrect relative order. And adding more rings won't change that incorrect relative order.
A: $4^5$ is the correct answer, but to an altered version of the problem, namely this:

Given a predetermined sequence of five rings which must be put on in that order, how many ways are there of wearing them on four fingers?

In this problem, we have exactly one way of choosing each successive ring: we take the next item from the predetermined sequence.
Yet some configurations are not possible under this constraint. For instance, if we're required to put on the rings in ABCDE order, it means that we can never produce a configuration in which the E ring is covered by another ring which was put on after it. Since E is put on last, it is the outermost ring of whichever finger it goes to.
The original problem has no such constraint: the rings are chosen in any order. So, when we put on the first ring, we not only have a choice of four fingers, but of five rings. Then a choice from among four rings, three, two and one.
But not all the choices of rings produce unique configurations. Given different ring-wearing sequences like ABCDE and BACDE, we can still easily produce an identical configuration. Under both these orders we can, for instance, put A on the index finger, B on the middle finger, and CDE on the pinky. Permutation of the rings provides some extra configurations, but not in a trivial way.
It's actually fairly surprising that the result is nothing but a tidy rising product from 4 to 8.
