Stacking rings on fingers by decreasing stack size (permutations)

[archaic]

Homework Statement

Let $x$ be the numbers of ways that a woman can wear $5$ distinct rings on $5$ fingers of her right hand, given that she can stack maximum of $3$ rings on any finger.

Relevant Equations

N/A

When she is stacking $5$ rings, then there are $5P5$ configurations when it comes to arranging rings, and each configuration can be arranged on her fingers in $5P1$ ways (choose one finger from 5 to put that configuration on).

When she is stacking $4$ rings, then we have two objects; a configuration of $4$ rings, and a ring outside. The number of configurations is also given by $5P5$. For the fingers, however, we now choose two fingers out of five to put the objects on, or, $5P2$.

The same reasoning goes for the last case, so$$x=5P5(5P3+5P2+5P1)$$

EDIT: I understood the problem statement in a peculiar way. I don't know how that happened. However, the logic is kind of still the same, and I think$$x=5P5(5P5+5P4+5P3)$$

 

[archaic]

I didn't consider the cases where we can stack 2 rings on 2 fingers, for example, etc ..

To make it easier on myself, I can first ignore how many rings are making a specific stack, and just consider objects to be arranged on the five fingers. For example, I'll just look at a configuration of $2, 2, 1$ as just $3$ objects to be arranged.

The number of objects I can possible have is $5, 4, 3$ and $2$, so $x=5P5a+5P4b+5P3c+5P2d$, where those variables need to be determined.

When we have $5$ objects, this means that I have a ring per finger, so $a=1$.

When we have $4$ objects, this means that there is a stack of $2$ and $3$ single rings. The number of ways I can gather a stack of $2$ is $5P2$, so $b=5P2$.

When we have $3$ objects, then we either have a stack of $3$ rings, and $2$ single rings, or $2$ stacks of $2$ rings, and single ring.
For the case where we have a stack of $3$, then this is similar to the above case, i.e we have $5P3$ ways of arranging that stack.
Basically, I can represent one configuration of the set of objects to be permuted by $\{(r_ar_b),(r_cr_d),r_e\}$. At first, I have $5P2$ ways of choosing $(r_ar_b)$, then I'd be left with $3$ rings, so I can choose $(r_cr_d)$ in $3P2$ ways. Moreover, I'll be doing twice the same permutation eventually since I'd have to permute $\{(r_ar_b),(r_cr_d),r_e\}$ and $\{(r_cr_d),(r_ar_b),r_e\}$.
Finally, I get $c=5P3+\frac{5P2\times3P2}{2}$.

When we have $2$ objects, then we either have $4,1$, or $3, 2$.
The first case is again like the others, so $5P4$ permutations.
For $3, 2$, it is like the last case. We have $\{(r_ar_br_c), (r_dr_e)\}$, so $5P3\times2P2$.
Thus $d=5P4+5P3\times2P2$.