Let $x$ be the number of ways that a woman can wear $5$ distinct rings on the $5$ fingers of her right hand, given that she can stack a maximum of $3$ rings on any finger. Find $x$.
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 possibly 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.
Finally, I get $c=5P3+5P2\times3P2$.
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 $d=5P4+5P3\times2P2$.
Best Answer
We multiply the number of ways of arranging the rings by the number of ways of distributing them to the fingers of the woman's right hand.
There are $5!$ ways to arrange five distinct rings. We place them on the fingers of the right hand from the bottom ring on the thumb (or the first finger on which a ring appears) to the top ring on the pinky (or the last finger on which a ring appears).
Next, we count the number of admissible ways of distributing the five rings to the fingers of her right hand.
Let $x_1$ be the number of rings placed on the thumb, $x_2$ be the number of rings on the index finger, $x_3$ be the number of rings placed on the middle finger, $x_4$ be the number of rings placed on the ring finger, and $x_5$ be the number of rings placed on the pinky. Since there are a total of five rings $$x_1 + x_2 + x_3 + x_4 + x_5 = 5 \tag{1}$$ Equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of $5 - 1 = 4$ addition signs in a row of five ones. For instance $$+ 1 + 1 + 11 + 1$$ corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 1$, $x_4 = 2$, $x_5 = 1$. The number of solutions to equation 1 in the nonnegative integers is $$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$ since we must choose which four of the nine positions required for five ones and four addition signs will be filled with addition signs.
However, we have the restriction that at most three rings may be placed on any one finger. Notice that at most one finger could have more than three rings on it since $2 \cdot 4 = 8 > 5$. There are five ways to select the finger which has more than three rings on it. Suppose that finger is the pinky. Then $x_5' = x_5 - 4$ is a nonnegative integer. Substituting $x_5' + 4$ for $x_5$ in equation 1 yields \begin{align*} x_1 + x_2 + x_3 + x_4 + x_5' + 4 & = 5\\ x_1 + x_2 + x_3 + x_4 + x_5' & = 1 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers with five solutions. Hence, there are $$\binom{5}{1}\binom{1 + 5 - 1}{5 - 1} = \binom{5}{1}\binom{5}{4}$$ solutions of equation 1 that violate the restriction that at most three rings may be placed on one finger.
Therefore, there are $$5!\left[\binom{9}{5} - \binom{5}{1}\binom{5}{4}\right]$$ ways to distribute five distinct rings to the five fingers of her right hand so that a maximum of three rings is placed on any finger.
Notice that it is only necessary to arrange the rings once. This is why your answer is too large.