If there are three fingers and 8 rings, how many ways of putting rings on fingers? 
This is a variation of the problem in the link below, while the original assumed that you could put more than one rings on one finger. I was wondering if we were to have 8 different types of rings and 3 fingers, how many ways can the rings be placed on the fingers if there are no more than 1 ring per finger?

I have a proposed idea on how to solve this problem though I doubt if my approach is correct. Unlike the original problem, the stars and bars method won't work. We have to make the problem divided into mutually exclusive cases, and I was thinking of doing the following:
Case 1: 1 ring ${8 \choose 1}{3 \choose 1}$
Case 2: 2 rings ${8 \choose 2}{3 \choose 2}$
Case 3: 3 rings: ${8 \choose 3}3!$
But my real trouble was not the ${8 \choose x}$, where $x$ is the number of rings, rather it was ${3 \choose y}$ where $y$ is the number of fingers. To clarify, I have doubts because I was wondering that I was going to have to multiply by $3!$ for every single one instead because there were 6 ways of arranging three fingers?
The Original Problem
 A: You are very close. There are indeed three cases:


*

*One ring: choose 1 ring out of 8 and 1 finger out of 3, with $1!$ ways in which the rings can be placed, for a total of ${8 \choose 1} \cdot {3 \choose 1} \cdot 1! = 24$ possibilities.

*Two rings: choose 2 rings out of 8 and 2 fingers out of 3, with $2!$ ways in which the rings can be placed, for a total of ${8 \choose 2} \cdot {3 \choose 2} \cdot 2! = 168$ possibilities.

*Three rings: choose 3 rings out of 8 and 3 fingers out of 3, with $3!$ ways in which the rings can be placed, for a total of ${8 \choose 3} \cdot {3 \choose 3} \cdot 3! = 336$ possibilities.
In total, there are $24 + 168 + 336 = 528$ possibilities. Note that if it is also allowed to select no rings at all, there are $529$ possible scenarios.
A: Pretty close, but you have to count the same things each case. (And include the fourth case of: ways to put zero rings on the fingers.)
It is:$$\require{cancel}{\quad\cancelto{1}{\binom 80\binom 30~0!}+\binom 8 1 \binom 3 1~ 1!+\binom 82 \binom 32 ~2! +\binom 83\binom 33 ~3!~\\=~529}$$
Counting for selecting 0,1,2, or 3 from 8 rings, selecting that number from 3 fingers, and arrange those selected rings on those selected fingers.
