How many ways can $5$ rings be placed on $4$ fingers?

Solution 1:

In the first case, if you are assuming all five rings are worn, you're close; you just mixed up your $n=4$ and $k= 5$. For the first: we can look at this as the number of ways to sum 4 non-negative integers to equal $5$: $$\underbrace{f_1 + f_2 + f_3 + f_4}_{n = 4\;\text{fingers}} = \underbrace{5}_{5\; \text{rings}}$$ where the $f$'s stand for fingers (four of them), and the 5 is the number of rings. This gives us: $$\binom{4 + 5 -1}{5} = \binom{8}{5} = \dfrac{8!}{5!3!} = \dfrac{8\cdot 7\cdot 6}{3\cdot 2 \cdot 1} = 56\;\;\text{combinations}$$

In the second case, there is no assumptions being made by your method: counting every possible way of wearing no ring, one ring...5 rings, on any one or more fingers gives us $5^4$ ways of wearing distinct rings, or no ring at all.

If you want to require that all rings be worn, as in case one, then we multiply the result of $(1)$ by $k! = 5!$, which gives all permutations of the rings, as they are distinct, and can thus be rearranged: that gives us $$(5!)\binom{4 + 5 -1}{5} = (5!)\binom{8}{5} = (5!)\dfrac{8!}{5!3!} = (5!)\dfrac{8\cdot 7\cdot 6}{3\cdot 2 \cdot 1} = (5!)56\;\;$$

Solution 2:

This sounds like the classic stars and bars problem, and your formula is wrong.

You need to put 5 rings on 4 fingers. So consider a string of 5 R's and 3 bars, giving you such a picture. E.g. RRR||R|R says put $(3,0,1,1)$ rings on fingers $(1,2,3,4)$. It's easy to convince yourself there is an exact correspondence between such strings and arrangements that you are seeking. So it's enough to count the number of such strings.

But in such a string, there are $n=5$ places for R and $k-1=3$ places for bars, and picking either all stars or all bars should be enough (because all other places will be filled by the other symbol). Since there is a total of $n+k-1 = 8$ positions, there is a total of $$ \binom{n+k-1}{n} = \binom{8}{5} = \frac{8!}{5!\cdot 3!} = 56 $$ such strings.

Alternatively, picking bars, you get $$ \binom{n+k-1}{k-1} = \binom{8}{3} = \frac{8!}{5!\cdot 3!} = 56 $$

as well.

Solution 3:

By the stars-and-bars approach, the solution for part 1 is $n+k-1\choose n$. For the second part note that different ring orders per finger may produce different outcomes. This is accounted for by simply multiplying the part 1 result with $n!$ (write down that stras-and-bars again and note that the stars can be permuted arbitrarily.

The number $\lim_{n\to\infty} n!{n+k-1\choose k}$ might be called Liberace constant.