Number of ways combination By how many dieffernet way can you place $3$ identical balls in $4$ boxes? 
I used  $\binom{n+r-1}n$ =$\binom{6}4=15$ . Is it right?
As there is a fair bit of confusion in some of the comments and in the posted solutions, let me write out an answer.  Throughout I will assume that the boxes are distinguishable.  

First of all, a direct count:  There are only three ways to write $3$ as the sum of positive integers $3=3,\;3=2+1,\;3=1+1+1$  Each gives us a pattern for your box fills.  

$3=3$, for example, gives us $\fbox 4$ solutions:  $\{3,0,0,0\},\;\{0,3,0,0\},\;\{0,0,3,0\},\{0,0,0,3\}$.  A quicker way to count is to simply remark that the solution is determined by specifying which box gets all the balls and there are $4$ choices. 

$3=2+1$ gives us $\fbox {12}$.  The pattern is $\{2,1,0,0\}$ and we have $4$ places to put the $2$ and then $3$ places to put the   $1$, and $4\times 3=12$

$3=1+1+1$ gives us $\fbox 4$ as there are $4$ places to put the $0$.  

Combining these we get $$4+12+4=\fbox {20}$$

Second method: Stars and Bars.  This is the formula you are trying to use, but mixing up the variables. To be sure, for larger collections a direct count is impractical so it's good to learn the general technique. To do it properly, write the balls as three stars in a line $$\star\star\star$$  We now insert three bars in the spaces between them and on either side, for example $$\star / \star // \star$$  Interpreting the bars as the gaps between the boxes we see that the indicated pattern corresponds to the solution $\{1,1,0,1\}$. Similarly $$//\star\star/\star$$ corresponds to the solution $\{0,0,2,1\}$ and $$\star\star\star///$$ gives us $\{3,0,0,0\}$.

In this way we see that the set of solutions is in bijection with the ways to place the three stars in six slots.  Hence the answer is $$\binom 63=\fbox {20}$$
I refer you to "Number of ways of distributing n identical objects among r groups" also in the Mathematics section of Stack Exchange.  Looks like you are almost there; the answer is {n+r /choose r} = 35 in your case, if I have understood your question properly. (NB The order in which the balls are placed is not accounted for in a combination calculation, so there are quite a few ways of distributing the balls as, e.g., 2 in 1 box and 1 in another here.)