Finding the number of solutions I am trying to compare my answer with a friend's and we are both confident in our answers. But the problem is, they are different. So the problem goes:
Suppose I have the equation 
$$x+y+z+w = 14$$
where $x,y,z,w \in \mbox{Z}^+$ (i.e. integers that can be zero or positive) such that $x,y,w,z \le 6$.
Friend's approach:
let $X=x+1, Y=y+1, Z=z+1, W=w+1$, so
$$X+Y+Z+W = 10$$
and without the final restriction, applying stars and bars, we get the number of ways :
$$ \left( \begin{array}{cc}
13  \\
3  \end{array} \right)$$
but we need to exclude cases where $1$ of the $4$ has $6$ or more, thus the final answer is:
$$ \left( \begin{array}{cc}
13  \\
3  \end{array} \right)- 
\left( \begin{array}{cc}
4  \\
1  \end{array} \right)\left( \begin{array}{cc}
7  \\
3  \end{array} \right)=146$$
My Way:
I incorporated the inclusion-exclusion idea
So using: $$ \left( \begin{array}{cc}
r+M-1  \\
M-1   \\ \end{array} \right)$$ where M represents different kinds of objects and r is the number of such objects, 
I first found all of the possible nonnegative solutions with no restrictions:
$$ \left( \begin{array}{cc}
14+4-1  \\
4-1   \\ \end{array} \right) =  \left( \begin{array}{cc}
17  \\
3   \\ \end{array} \right)$$
Then I let $x,y,z,w \ge 7$ to be used for the solutions that are not allowed and since there are four variables and their restrictions are all the same, I multiply the following by $C(4,1) = 4$:
$$ 4\left( \begin{array}{cc}
(14-7)+4-1  \\
4-1   \\ \end{array} \right) =  4\left( \begin{array}{cc}
10  \\
3   \\ \end{array} \right)$$
And for the last part, I kept the restriction but now choose two variables of the four. I'll also be multiplying the following by $C(4,2) =6$ since there are six ways of pairing up the variables so:
$$ 6\left( \begin{array}{cc}
(14-7-7)+4-1  \\
4-1   \\ \end{array} \right)=  6\left( \begin{array}{cc}
3  \\
3   \\ \end{array} \right)$$
So by the inclusion-exclusion idea, the number of solutions:
$$\left( \begin{array}{cc}
17  \\
3   \\ \end{array} \right)- 4\left( \begin{array}{cc}
10  \\
3   \\ \end{array} \right) + 
6\left( \begin{array}{cc}
3  \\
3   \\ \end{array} \right) = 206$$
Conclusion:
So which one of us is correct? I've asked this problem before and had a member from this site confirm my answer already so I am set on it. But my friend has two confirmations from math/cs majors. Any help is greatly appreciated. 

Your friend is counting the solutions where $ 1 \le w,x,y,z \le 6$; you are counting the solutions where $ 0 \le w,x,y,z \le 6$.
Since the problem statement says $x, y, z, w \in Z^+$, I think your friend has the proper interpretation.
