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 $
 A: From your attempt we know that this problem is equivalent to the number of non-negative integer solutions to $m + n + o + p = 11$, which is far simpler.
This type of problem is sometimes known as "Stars and Bars" http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29. The idea is that you have 11 "stars", how many different ways can you put 3 bars in between them to separate them into 4 groups? In total, you have $11 + 4 - 1 = 14$ spaces for either a star or a bar, from that you need to choose $4 - 1 = 3$ spaces to place a bar. This equates to $\binom{14}{3} = 364$.
A: Hint:
Find the coefficient of $x^{15}$ of the following function: 
$$f(x)=(x+x^2+x^3+\cdots+x^{15})^4$$
Why this gives you the number of integer solution of your equation?
A: 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 
$$N(x+y+z+w=15)=\sum_{n=1}^{12}n*(13-n)=364.$$
A: There is a formula for the number of solutions in $N^p$ of the equation $x_1+ ... +x_p=n$, notably $C_{n+p-1}^{p-1}$. 
A: A visual solution I learned from my teacher was using spaces (to make it easier to understand the math). Since the total is 15, we have 15 spaces. BUT, because we are using positive numbers (not non-negative), then we have to add 1 more to each variable (to make it positive if it were to be 0), which means we subtract how much we added from the total. So we have 11 spaces.
_ _ _ _ _ _ _ _ _ _ _ 
Since we have 4 variables, we add (4-1) zeros to separate the spaces into 4 sections.
_ _ 0 _ _ 0 _ _ 0 _ _
Since the zeros took 3 spaces, we add 3 spaces wherever you want.
_ _ _ 0 _ _ 0 _ _ _ 0 _ _ _
Notice that for the above spaces, we have 3 + 2 + 3 + 3 = 11. That is 1 solution, but we can't keep doing that. It's not efficient. So here we can use the total number of spaces, 14, and partition it into 4 sections, so we have:
(p + n - 1) choose (n - 1) => (11 + 4 - 1) choose (4 - 1) => (14 choose 3) = 364.
Where p is the total of the equation and n is the number of variables. Notice that the above equation is the visual I used. Same thing.
