<s>[INST] How many Positive integer solutions does the equation $x + y + z + w = 15$ have? How many Positive integer solutions does the equation $x + y + z + w = 15$ have?
Attempt:
Let $x = m + 1, y = n + 1, z = o + 1, w = p + 1 $
Then, $ m + 1 +  n + 1 + o + 1 +  p + 1 = 15$
$ m + n + o + p = 11 $
 [/INST] Here is my try. 
Your equation is $x+y+z+w=(x+y)+(z+w)=15$. First we see $x+y$ and $z+w$ as two unknowns, that is $a+b=15$ and $a,b$ satisfy $2\leq a,b\leq13$. Easily, we can say that there are $12$ positive integer solutions for $a$ and $b$. Then we will see there are how many postive integer solutions for $x+y=a$ and $w+z=b$. We note the number of such solutions as $N(\cdot)$.
If $a=2$, then $b=13$. We see that $a=2=x+y$ has unique $1$ solutions for $x$ and $y$, that is $x=1$ and $y=1$. $b=13=z+w$ has $12$ solutions for $z$ and $w$, that is $z=1,2,\dots,12$ and $w=12,11,\dots,1$. Then there are $N(a=2)*N(b=13)=1*12=12$ solutions for $a=2$ and $b=13$.
Then we do like this, we can make a list of the 12 solutions for $a$ and $b$:
$$N(a=2)=1\Leftrightarrow N(b=13)=12\Leftrightarrow N(a=2,b=13)=1*12$$
$$N(a=3)=2\Leftrightarrow N(b=12)=11\Leftrightarrow N(a=3,b=12)=2*11$$
$$\vdots$$
$$N(a=13)=12\Leftrightarrow N(b=2)=1\Leftrightarrow N(a=3,b=12)=12*1$$
So the number of all the solutions is 
$$N(x+y+z+w=15)=\sum_{n=1}^{12}n*(13-n)=364.$$
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