Assume each ring is unique, in how many ways can you wear 6 rings on 4 fingers? No restrictions, so 1 finger can have all 6 rings.

My solution so far:
There are 4 ways to get 6 rings on one finger, and each of those has 6P6 ways to be permutated.
4*3 ways to get 5 rings on one and 1 ring on another, 6P5*1P1 permutations
4*3 ways to get 4 on one 2 on another, 6P4*2P2 permutations
4*3 ways to get 4 on one 1 on another and 1 on third, 6P4*2P1*1P1 permutations
6 ways to get 3 on one 3 on another, 6P3*3P3 permutations
4*3 ways to get 3 on one, 2 on another, 1 on third, 6P3*3P2*1P1 permutations
4 ways to get 3 on one, 1 on the rest, 6P3*3P1*2P1*1P1 permutations
4 ways to get 2 on three, 6P2*4P2*2P2 permutations
6 ways to get 2 on two, 1 on the other two, 6P2*4P2*2P1*1P1 permutations

Excel gives 51,840

is my approach correct? Is there a simple generalization to R unique objects stacked in N groups?

This is a well known problem which states that there are C(n+r -1, r-1) distinct number of nonnegative integer valued vectors (x1,x2,...,xr) satisfying x1 + x2 + ... + xr = n with "disregard to order". The formula where order matters (your example) is n!*C(n+r-1,r-1).

Plugging in the numbers where n = 6 and r = 4 gives us 60,480

Lets work this out in cases:

n(6,0,0,0) = C(4,1)*P(6,6)

n(5,1,0,0) = C(4,2)*2*P(6,5)*P(1,1)
n(4,2,0,0) = C(4,2)*2*P(6,5)*P(1,1)
n(3,3,0,0) = C(4,2)*2*P(6,5)*P(1,1)

n(4,1,1,0) = C(4,3)*3*P(6,4)*P(2,1)*P(1,1)
n(3,2,1,0) = C(4,3)*6*P(6,3)*P(3,2)*P(1,1)
n(2,2,2,0) = C(4,3)*1*P(6,2)*P(4,2)*P(2,2)

n(3,1,1,1) = C(4,1)*P(6,3)*P(3,1)*P(2,1)*P(1,1)
n(2,2,1,1) = C(4,2)*P(6,2)*P(4,2)*P(2,1)*P(1,1)

Add everything up and we get 60,480

Also notice that when you multiple the P's out, you always get 6! = 720 which becomes a common factor in the total sum. I wrote out the necessary steps so that it's easier to understand.

This problem has come up a few times here. I've given an outline on how to derive the formula which is given in the thread below.

http://forumserver.twoplustwo.com/sh....php?p=2437170

Thanks for your help.

Drill nine holes in a row. Starting from the left, either put one ring in a hole or leave it empty. Put your pinky in the first (of three) empty holes again starting from the left. All rings to the left of that pinky, if any, go on it. Moving to the right until you get to the next empty hole, put your ring finger in. Put on it the unused rings to its left, if any. Do the same for your middle finger. If your middle finger isn't in the far right hole, put the ring in that hole and all the (unused) rings immediately adjacent too it on your index finger.

Any combination of six identical rings on four fingers is obviously in one to one correspondence to putting six rings in nine holes. So there is nine choose six ways of doing that. 84. If the rings are different they can be ordered in 720 ways for each combination. 60480.

I didnt fully understand it until I read the guide from
http://www.mathsisfun.com/combinator...mutations.html

The topic about Combinations with Repetition
is the example to six identical rings

for unique rings, then you multiply your above result with 6! = 720