First check out this excellent post: https://www.beatthegmat.com/a/2009/10/12 ... asy-method. Simplest explanation of permutations and combinations I've ever seen.

Now, an important caveat he mentions is that it doesn't apply to situations where you can use the item/person more than once.

How would you modify the methods in this case? Suppose you had five colors of stones, and you want to set three stones in a ring. Assuming you have unlimited of each color, you could determine this by simply not subtracting one for each.

ONE OF EACH COLOR
5 x 4 x 3 = 60

UNLIMITED OF EACH COLOR
5 x 5 x 5 = 125 ways to set three stones in the ring, where each way is unique in the amount and position of each color.

However, what if the question was how many different combinations of color are there?

ONE OF EACH COLOR
[5 x 4 x 3] / [1 x 2 x 3] = 60 / 6 = 10

UNLIMITED OF EACH COLOR
[5 x 5 x 5] / [1 x 2 x 3] = 125 / 6 = Not integer?

I assume you can't take 125 / 6, because it's not an integer, so there needs to be a way to modify the combination approach. Any ideas?

Update: Maybe there's not a way to modify? I listed out all combinations, and I got 35.

C1 = Color 1, C2 = Color 2, etc.

All C1
All C2
All C3
All C4
All C5

Total = 5

2 C1, other color (4)
2 C2, other color (4)
...

Total = 20

C1, C2, C3
C1, C2, C4
C1, C2, C5
C1, C3, C4
C1, C3, C5
C1, C4, C5
C2, C3, C4
C2, C3, C5
C2, C4, C5
C3, C4, C5

Total = 10

Total combinations = 35

35 is not a factor of 125, so doesn't appear to be an easy way to calculate combinations with replacement.