Hello, I came across a combinatorics problem that has raised some troubling issues for me and was wondering if anyone could clarify.

How many ways are there to place nine different rings on the four fingers of your right hand (excluding the thumb) if:

a) The order of rings on a finger does not matter?

b) The order of rings on a finger is considered?

Now the answer to part a) is $\displaystyle 4^9$

For each of the nine different rings, we have four choices of fingers.

The answer to part b) is $\displaystyle 9! * \binom{9+4-1}{9}$

First we see how many ways we can distribute the rings across our four fingers, then we order them.

What I am wondering is why isn't $\displaystyle \binom{9+4-1}{9}$ the answer to part a) of the question?

Is the calculation $\displaystyle \binom{9+4-1}{9}$ treating the rings as if they were indistinct?

And it is just easier to treat them as indistinct first to order them for part b)?

Any clarification would be greatly appreciated.