Combining four distinct objects with repetition 
With four different objects $k = \{obj1, obj2, obj3, obj4\}$. How many combinations are there if I were to copy $20$ freely-chosen objects. I could for example have $20\times obj1$ if I wanted.
However, permutations are not to be counted.

Probable solution
I'm thinking I need to use one of the following (theorems?): $$\binom{n+k-1}{k}\quad or\quad \binom{n+k-1}{n}$$ However, there are two problems with this. I wouldn't know which one, and I can't find any understandable explanation. The latter I withdrew from the theorem from stars and bars method.
What's the difference betweeen the formulas? And should I even use one of these formulas for my problem?
The first yields $8855$ combinations, and the second yields $1771$ combinations.
Further question
Wherever I look, people use the formulas interchangeably$^{[1]}$, however; they're not equal. Take for example: $$\binom{9+3-1}{3} \not= \binom{9+3-1}{9}$$
How come?
$^{[1]}$ http://mathworld.wolfram.com/Multichoose.html
http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29#Theorem_two_2
 A: The number of solutions of $x_1+x_2+\cdots+x_k=n$ in non-negative integers is
$$\binom{n+k-1}{k-1}\quad\text{or alternately}\quad\binom{n+k-1}{n}.$$
In your further question, it looks as if you are using $n=9$ and $k=3$. Then the first version should be $\binom{9+3-1}{3-1}$, not $\binom{9+3-1}{3}$. 
In the original question, we want to find the number of solutions in non-negative integers of the equation
$$x_1+x_2+x_3+x_4=20.$$
By standard Stars and Bars, the number is $\binom{20+4-1}{4-1}$, or equivalently $\binom{20+4-1}{20}$. 
Remark: The equivalence of the two versions follows from the fact that in general $\binom{m}{r}=\binom{m}{m-r}$. 
A: I'd break the cases down by the number of distinct objects in your set of $20$.
If you have (exactly) one distinct object, there are $4$ combinations (twenty of obj1, obj2, obj3, or obj4).
If you have (exactly) two distinct objects, you can divide them up $(19, 1), (18, 2), (17, 3), ... (10, 10)$.  Let's first look at the case $(x,y)$ where $10 < x < 20$.  You choose the object $x$, then the different object $y$.  There are $12$ ways to do this, and $9$ values of $x$, so there are $108$ combinations.  For $x=y=10$, there are ${4 \choose 2} = 6$ combinations.  So, the number of combinations for two distinct types of objects is $108+6=114$.
I'd build up the cases for three and four distinct objects in a similar way.
