Number of ways of arranging $5$ distinct rings on $5$ fingers As a task after Combinatorics, I was asked to calculate the number of ways of arranging $5$ distinct rings on $5$ fingers (counting the thumb as a finger). The emphasis on arranging is indicating that the order of which ring was put on in what order matters for a particular finger. Note that all $5$ rings are to be used in the arrangement.
Here's what I attempted. I counted few cases like those of putting on all $5$ in $1$ finger, all $5$ rings in selected $2$ fingers and so on.
$$\text{ways}=\underbrace{{5\choose 1 }5!}_{\text{all 5 in 1}}+\underbrace{\left[{5\choose 2}2!\left({5\choose 1}4!+{5\choose 2}2!3!\right)\right]}_{\text{all 5 in 2}}+\underbrace{\left[{5\choose 3}\left\{{5\choose 1}{4\choose 1}3!+{5\choose 1}{4\choose 2}2!2!+{5\choose 2}{3\choose 1}2!2!+{5\choose 2}{3\choose 2}2!2!+{5\choose 3}{2\choose 1}3!+{5\choose 1}{4\choose 3}3!\right\}\right]}_{\text{all 5 in 3}}+\underbrace{\left[{5\choose 4}\left\{{5\choose 1}{4\choose 1}{3\choose 1}2!+{5\choose 1}{4\choose 1}{3\choose 2}2!+{5\choose 1}{4\choose 2}{2\choose 1}2!+{5\choose 2}{3\choose 1}{2\choose 1}2!\right\}\right]}_{\text{all 5 in 4}}+\underbrace{5!}_{\text{all 5 in 5 }}$$
All this evaluates to $15120$. Could you check my work? Also suggestions to shorten the approach are welcome. Thanks

Edit $1$: I've corrected the mistake I was making. The answer now matches with the expected answer. Thanks @N. F. Taussig for taking the time to explain what the discrepancy was.

Edit $2$: I also came to know about another method. So for the first ring say $\text{R}_1$ there are $5$ places, for the second ring there's $6$ places, the $4$ remaining fingers and for the case of that finger being the same on which we put the first ring on there's $2$ options, either above it or below it, and since the job of placing the second ring would be said to be completed in either way the addition principle applies, thus amounting to $6$ possible places. So forth and so on till we've accounted for all the rings. This gives us the number of ways as $5\times 6\times 7\times 8\times 9=9!/4!={9\choose 4}\times 5!$.

Method 1:  We choose how many rings to place on each of the five fingers, then multiply by the $5!$ of arranging the rings from bottom left to top right.  Let $x_i$ be the number of rings placed on the $i$th finger.  Then
$$x_1 + x_2 + x_3 + x_4 + x_5 = 5$$
is an equation in the nonnegative integers.  A particular solution of the equation corresponds to the placement of four addition signs in a row of five ones.  For instance,
$$+ 1 + 1 + 1 1 + 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 1$, $x_4 = 2$, $x_5 = 1$ (no rings on the thumb, one ring each on the index, middle, and pinky fingers, and two rings on the ring finger).  The number of such solutions 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. Multiplying by the $5!$ ways of arranging the rings from bottom left to top right yields
$$\binom{9}{4}5!$$
Method 2:  We arrange five distinct rings and four dividers, which correspond to the decision to jump from one finger to the next, in a row.  Choose four of the nine positions for the dividers, then arrange the five distinct rings in the remaining five positions.  Again, we obtain
$$\binom{9}{4}5!$$
ways of arranging five distinct rings on five fingers.
Where did you make a mistake?
Your calculations for distributing five distinct rings to one finger, two fingers, and five fingers are correct.
I do not follow your calculations for three fingers or four fingers.
Distributing five rings to three fingers:  Either one of the three selected fingers receives three rings with the others each receiving one or two of the three selected fingers each receive two rings with the other receiving one.
One of the three selected fingers receives three rings with the others each receiving one:  There are $\binom{5}{3}$ ways to select which three fingers receive a ring, $\binom{3}{1}$ ways to select which of those fingers receives three rings, $\binom{5}{3}$ ways to select three rings for that finger, $3!$ ways to arrange the rings on that finger, and $2!$ ways to arrange the remaining two rings on the remaining two fingers, giving 
$$\binom{5}{3}\binom{3}{1}\binom{5}{3}3!2!$$
such distributions.
Two of the three selected fingers each receive two rings with the other receiving one:  There are $\binom{5}{3}$ ways to select which three fingers receive a ring, $\binom{3}{2}$ ways to select which two of those fingers receive two rings, $\binom{5}{2}$ ways to choose which two rings are placed on the leftmost of those fingers, $2!$ ways to arrange the ring on that finger, $\binom{3}{2}$ ways to choose which two the remaining rings are placed on the other finger that receives two rings, $2!$ ways to arrange those rings on that finger, and one way to place the remaining ring on the remaining finger.  There are 
$$\binom{5}{3}\binom{3}{2}\binom{5}{2}\binom{3}{2}2!2!$$
such distributions.
Therefore, the number of ways to distributing five distinct rings so that  exactly three of the five fingers receive a ring is
$$\binom{5}{3}\binom{3}{1}\binom{5}{3}3!2! + \binom{5}{3}\binom{3}{2}\binom{5}{2}\binom{3}{2}2!2!$$
Distributing five rings to four fingers:  It must be the case that one of the four selected fingers receives two rings, while each of the others receive one ring.  There are $\binom{5}{4}$ ways to select the fingers that receive a ring, $\binom{4}{1}$ ways to choose which of those fingers receives two rings, $\binom{5}{2}$ ways to choose which two rings are placed on that finger, $2!$ ways to arrange those rings on that finger, and $3!$ ways to arrange the remaining three rings on the remaining three fingers.  There are 
$$\binom{5}{4}\binom{4}{1}\binom{5}{2}2!3!$$
such distributions.
With these corrections, you will find that there are $15,120$ ways to distribute five distinct rings to five fingers when the order of the rings on each finger matters, in agreement with the answer we obtained above. 
