Cups of water from a bucket We have an empty container and $n$ cups of water and $m$ empty cups. Suppose we want to find out how many ways we can add the cups of water to the bucket and remove them with the empty cups. You can use each cup once but the cups are unique. 
The question:  In how many ways can you perform this operation.
Example: Let's take $n = 3$ and $m = 2$.
For the first step we can only add water to the bucket so we have 3 choices.
For the second step we can both add another cup or remove a cup of water.
So for the first 2 steps we have $3\times(5-1) = 12$ possibilities.
For the third step it gets more difficult because this step depends on the previous step.
There are two scenarios after the second step. 1: The bucket either contains 2 cups of water or 2: The bucket contains no water at all.
1) We can both add or subtract a cup of water
2) We have to add a cup of water
So after step 3 we have $3\times(2\cdot3 + 2\cdot2) = 30$ combinations.
etc.
I hope I stated this question clearly enough since this is my first post. This is not a homework assignment just personal curiosity. 
 A: Let $L=n+m$ and $D=n-m \ge 0$
Hint 1: consider a sequence $X=(x_1,x_2, ... x_L)$, associate a filled cup with $x_i=1$ and an empty with $x_i=-1$. Let $C(n,m)$ count all such binary sequences of length $L$ with the two restrictions: $\sum_{k=1}^j x_k\ge 0$, $\forall j$ and $\sum_{k=1}^L x_k =D$
Then, the total number of ways is $C(n,m) n! m!$
Hint 2: To obtain $C(n,m)$ is not trivial, but random walks counting procedures (mirror principle) might help (eg).

Edited: using the reflection principle: 
Consider first counting all the paths going from $y(0)=0$ to $y(L)=D$  such that $y(t+1)=y(t) \pm 1$, and with $D\ge0$. (Recall that $L=n+m$, $D=n-m$). This unrestricted count is $C^{[u]}={n+m \choose n}={L \choose (L+D)/2}$
Now, consider the "prohibited" paths: these correspond to those that touch the $y(t)=-1$  line. By the mirror principle, these correspond one-to-one to the unrestricted paths that start from $y(0)=-2$, and this count is given by  $C^{[p]}={L \choose (L+D+2)/2}={n+m \choose n+1}$
Hence, the number of allowed paths is
$$C(n,m) = C^{[u]}-C^{[p]}={n+m \choose n} - {n+m \choose n+1}=\frac{1+n-m}{n+1}{n+m \choose n} = \frac{1+D}{n+1}{L \choose n}  $$ 
And, finally, the number of ways is
$ \frac{1+D}{n+1} \,L!$
See also the Ballot problem.
A: For $n=m$ the answer is $C_n(n!)^2$ where $C_n=\frac 1{n+1}{2n \choose n}={2n \choose n}-{2n \choose n-1}$ is the $n^{\text{th}}$ Catalan number.  The walk on the chessboard staying below the diagonal matches your constraint of never having a negative amount of water in the bucket and the two factors of $n!$ come from the labeled cups.  For $n \gt m$ I want to imagine the additional $n-m$ empty cups added and requiring them to be at the end of the series, but can't see how to account for that.
