How many integer solutions are there for the equation $c_1 + c_2 + c_3 + c_4 = 25$, where $c_i \ge 0$ for all $1 \le i \le 4$ Question Statement:
How many integer solutions are there for the equation $c_1 + c_2 + c_3 + c_4 = 25$, where $c_i \ge 0$ for all $1 \le i \le 4$. 
I would like to solve this problem using combinatorics and I've read generating functions can be used as a method to find the solution. However, I have no idea how to do this.
My first attempt at solving this problem is below, 
Observe the missing constraint $c_i \le 21$. The solution can be obtained by reasoning using Principle of Exclusion and Inclusion.
Applying the theorem to the above problem yeilds, 
$N(\bar{c_1}\bar{c_2}\bar{c_3}\bar{c_4}) = N - \sum N(c_i) + \sum N(c_i c_j) - \sum N(c_i c_j c_k) + \sum N(c_1 c_2 c_3 c_4)$
For all $i,j,k = 1,...,4$. 
Since, $N=H(4,25)=C(28,25)$, $N(c_i)=H(4,4)=C(7,4)$ and $N(c_i c_j) = N(c_i c_j c_k) = N(c_1 c_2 c_3 c_4) = 0$. Hence, the result is 3248.  
 A: Generating functions is the hard way for this question, but here goes. 
The answer is the coefficient of $x^{25}$ in $(1+x+x^2+\cdots)^4$. We find $$(1+x+x^2+\cdots)^4=(1-x)^{-4}={3\choose0}+{4\choose1}x+{5\choose2}x^2+\cdots$$ so the answer is ${28\choose25}={28\choose3}=3276$. 
A: Generating Function Method
Associate to each variable the polynomial $p(x) = \sum_{i=0}^{25} x^i$.  Then the product
$$  \left(p(x)\right)^4 = 1 + 4 x + 10 x^2 + \cdots + 3276 x^{25} + \cdots  $$
exhibits the fact that there are $3276$ solutions to the equation.  It also exhibits the number of solutions of \begin{align*}
c_1 + c_2 + c_3 + c_4 &= 0 & :& & 1  \\
c_1 + c_2 + c_3 + c_4 &= 1 & :& & 4  \\
c_1 + c_2 + c_3 + c_4 &= 2 & :& & 10  \\
& & \vdots& &
\end{align*}
Our polynomial encodes the choices for the variable in the powers of $x$, so we have one term for each of the integers $0$ through $25$.  When you multiply two of these polynomials, you get generic terms $x^i x^j$ for $0\leq i,j \leq 25$.  But consider the terms we get for $i+j = 5$, for instance, they are 
$$  x^0 x^5, x^1 x^4, x^2 x^3, x^3 x^2, x^4 x^1, x^5 x^0, $$
that is, we have one term in the product for each way to write $5$ as a sum of two nonnegative integers, so the resulting product of two polynomials records the number of ways to produce $n$ as the sum of two nonnegative integers in the coefficient of $x^n$.  Multiplying in the other two polynomials, the coefficient of $x^n$ records the number of ways to write $n$ as the sum of four nonnegative integers (each less than $25$).
(One might wonder how to compute that massive product.  You don't, exactly.  You only need terms of degree up to $25$ throughout the computation, so you only keep track of the leading terms and ignore the rest.  For me, this computation went as \begin{align*}
p^1 &= 1 + x + x^2 + \cdots + x^{25} + \text{(don't care)} \\
p^2 &= 1 + 2x + 3x^2 + \cdots + 26 x^{25} + \text{(don't care)}  \\
p^3 &= 1 + 3x + 6x^2 + \cdots + 351 x^{25} + \text{(don't care)} \\
p^4 &= 1 + 4x + 10x^2 + \cdots + 3276 x^{25} + \text{(don't care)}
\end{align*}
It helped that I am familiar with figurate numbers and recognized the coefficients were, successively, constantly one, sequential integers, sequental triangular numbers, and sequential tetrahedral numbers.)
A: Two methods to solve this are "Stars and Bars" and Generating Functions.

Stars and Bars
we put $25$ $\star$s and $3$ $|$s to separate the $\star$s into $4$ areas. For example, $6+4+8+7$ would be represented by
$$
\overbrace{\star\star\star\star\star\,\star}^6|\overbrace{\star\star\star\,\star}^4|\overbrace{\star\star\star\star\star\star\star\,\star}^8|\overbrace{\star\star\star\star\star\star\star}^7
$$
each arrangement of the $25$ $\star$s and $3$ $|$s will give a unique sum. The number of such arrangements is
$$
\binom{28}{3}=3276
$$

Generating Functions
Each choice of $x_k$ so that $x_1+x_2+x_3+x_4=25$ corresponds to an $x^{25}$ term of
$$
\overbrace{\left(1+x+x^2+\cdots\right)}^{\frac1{1-x}}\overbrace{\left(1+x+x^2+\cdots\right)}^{\frac1{1-x}}\overbrace{\left(1+x+x^2+\cdots\right)}^{\frac1{1-x}}\overbrace{\left(1+x+x^2+\cdots\right)}^{\frac1{1-x}}
$$
and since
$$
\begin{align}
(1-x)^{-4}
&=\sum_{k=0}^\infty\binom{-4}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+3}{3}x^k
\end{align}
$$
the coefficient of $x^{25}$ is
$$
\binom{28}{3}=3276
$$
