How many ways can $8$ rings be put on $3$ fingers? I've been trying to understand an example question in my textbook but after looking around the answer seems unclear to me.

The question: 


  "How many combinations/permutations of $8$ different rings can be put on $3$ different fingers? Both if the order matters and also if it does not."


My attempt, if order does not matter would be :

$8^3$

If the order does matter:

$8P3 = 8 \cdot 7 \cdot 6$

Is this the correct way to interpret such a question? 

Some online resources seem to have varying and sometimes odd solutions such as this: $C(n+r -1, r-1)$

Any assistance in understanding if I'm heading to the correct interpretation would be helpful.

Edit

Note : More than $1$ ring can be put on each finger
For order doesn't matter, each ring can be placed on one of $3$ fingers. This results in a unique string, for example $12323113$, which results in finger $1$ having rings $1,6,7$, finger $2$ has rings $2,4$ and finger $3$ has rings $3,5,8$.

This clearly groups each finger's rings as distinct, and gives $3^8$ strings.

$8^3$ is three rings onto $8$ fingers.

If order does matter, using $3^8\cdot8!$ doesn't work, for example in the above case there are only $3!2!3!=72$ permutations.

Instead, use stars and bars to give the number of available patterns as $\dbinom{10}{2}=45$.

To feed the fingers their rings, first arrange the rings into one of the $8!$ permutations, and feed from finger $1$ through to finger $3$. This ensures uniqueness, and gives the result as:

$$\binom{10}{2}\cdot8!=45\cdot40320=1814400$$
If you're like me the problem is finding a strategy.   

And the strategy that figuring each ring has $3$ choices of fingers so there are $3^8$ ways to choose fingers for each ring, is a dead end as order matters and we can't straightforwardly multiply by any choice or permutation value for each finger as we don't know how many rings are on each finger.

We can choose to say each finger has $aPb = \frac {a!}{(a-b)!}$ where $a$ are the number of rings left, and $b$ are the number of rings on the finger:

Then the answer is $\sum_{k= 8,-1}^0 (8 P k)\sum_{j=0}^k(k P j)(k-j P k-j)$.  Which seems intimidating but: $(k P j)(k-j P k-j) = \frac {k!}{(k-j)!}(k-j)! = k!$ and so $sum_{j=0}^k (k P j)(k-j P k-j) = \sum k! = (k+1)k! = (k+1)!$ and $\sum_{k= 8,-1}^0 (8 P k)\sum_{j=0}^k(k P j)(k-j P k-j) = \sum_{k = 0}^8 \frac{8!}{k!}*(k+1)! = 8! \sum_0^8 (k+1) = 8!\sum_{i=1}^9 i = 8!\frac{9*10}2 = 8!*45$.

But simpler way of thinking of it is that if all the rings were identical and there were $N$ ways to place $8$ things on $3$ fingers.  And there are $8!$ ways to arrange the 8 rings.  There would be $8!N$ ways to arrange 8 different rings.  

There are two ways to solve $N$.  You may put $a$= zero to $8$ rings on finger $1$ and you may put $b$ = 0 to $8-a$ on finger 2 and all the rest on finger 3.  That is $\sum_{a=0}^8\sum_{b=0}^{8-a}1 = \sum_{a=0}^8 9-a = \sum{a=8,-1}^0 a+1 = \sum_{i=1}^9 i = \frac {9*10}2 = 45$.

Or: you need to divide $8$ onto three fingers. Put the on one after the other.  There are nine points in time between $0$ and $8$ that you can put in a "go to next finger marker".  Among the $0$ to $8$ ring and the marker there are 10 places to put a second marker.  There are ${10 \choose 2} =\frac {10!}{2!8!} = \frac {10*9}n = 45$.

Either way there are $45*8!$ ways to do this.

==== old answer which was stream of thought with mistakes along the way below === it could be illuminating for thought process... or it could be frustrating as it takes a while to get the right answer =====

If order doesn't matter then each ring can go on any one finger is $3^8$ (not $8^3$) is correct.

But if order on the fingers matters (the emerald on the middle finger under the gold is different than the emerald on the middle finger over the gold) it's a different question.

One way:  Arrange the rings in order that you will put them on.  There are $8*7*6= \frac {8!}{(8-3)!}$ (is that what $8 P 3$ defined to be?  I think so.) ways to arrange  .  Then each ring (in order) has a choice of three fingers to be placed on.  So the answer is $8P3*3^8$.

==== Oh, F###; that's over counting ====

If diamond ring is 1 and we choose finger one and emerald is 2 and we choose finger two is the same as emerald is 1 and we choose finger two and diamond is 2 and we choose finger one.

Oh, well, I'm going to leave this up as foood for thought. but... it is a wrong answer.

=====

My first thought was the hardest. You can choose $a$ rings for the first finger and $b$ rings for the second and $8 - a -b$ for the third and so

$\sum_{a= 0}^8 (8Pa)\sum_{b=0}^{8-b}([8-b]Pb)*([8-b-a]!)$

Logic tells me that sum must add up to $8P3*3^8$ (EDIT: It doesn't) but ... I'd hate to actually work it out (EDIT: but I may have to... it seems like the correct answer still.).

(Actually my very first thought was order didn't matter and rings were all the same, so it'd be $\sum_{a=0}^8 \sum_{b=0}^{8-a} 1 = \sum_{a=0}^8 9-a = 9^2 - \frac{8*9}4= 45$). 

====

Here's a second way.  We are going to put all the rings on in order all the rings on the first finger first, then the rings on the second, and then the third.

There are $8!$ ways to arrange there rings and $\sum_{a=0}^8 \sum_{b=0}^{8-a} 1 = 45$ ways to place the two "breaks" on when we stop putting rings on one finger and start putting them on the other.

So $8!*45$.  Is that right? 

Anyone with time on his/her hands want to see if $\sum_{a= 0}^8 (8Pa)\sum_{b=0}^{8-b}([8-b]Pb)*([8-b-a]!) = 8!*45=8! {10 \choose 2}$?