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.  

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
$$
\binom{28}{3}=3276
$$
