Number of solutions to $a+b+c+d=14$ Where $a,b,c,d\in\{0,1,2,\ldots\}$. I understand how to find to solution (now), however I'm asking why a particular method I tried failed to work.
I imagined laying out $14$ objects in a row, and placing $3$ dividers between two objects, or at the beginning or end. Then $a$ is the number of objects from the beginning up to the leftmost divider, then $b$ is the next up to the second leftmost divider, $c$ is the next up to the rightmost divider, and $d$ is all that remains. $a,b,c,d$ satisfy the problem, and all possible solutions to the problem are included.
There are $15$ slots to place a divider, and each divider can be placed independently of the other $2$. This gives $15^3$ ways to place the dividers assuming they are unique. But since they aren't, there are $15^3/3!$ ways to place the dividers, and thus there are those many solutions to the problem. However it is clear that this is not even an integer.
 A: You divide by $3!$ to compensate for the fact that permuting the dividers yields a different position of the dividers, but the same solution. But this is true only when the dividers are placed in different slots, if two or three dividers are placed in the same slot, you don't double count the solutions.
You need to count:


*

*Solutions with three dividers in three different slots (divide by $3!$).

*Solutions with two dividers in the same slot and the third in a different slot. Depending by the way to count these solutions, you might need to divide by $2!$ or not.

*Solutions with all three dividers in the same slot. 


In your solution you count all three cases at once, and divide by $3!$...
A: Your method will work, if you take into account that inserting a divider creates an additional gap.  
If you first treat the dividers as distinct, you have 15 choices for the position for the first divider, but then 16 choices for the second and 17 for the third; so the number of possibilities is given by $\hspace{2 in}\displaystyle\frac{15\cdot16\cdot17}{3!}=\binom{17}{3}$.
