Finding the coefficient in an expansion? I'm reviewing for my combinatorics final and have completely forgotten how to find the coefficient in the expansion of a polynomial. Here's an example I'm struggling with

Find the coefficient of $x^{11}$ in the expansion of
  $$(x+x^2+x^3+x^4+x^5)^7(1+x+x^2+x^3+\dots)^4$$

So first I know how to rewrite it as
$$x^7(1-x^5)^7\frac1{(1-x)^{11}}$$
But I have no idea how to find the coefficients from here.
 A: Hint Pulling out the factor of $x^7$ we see that this is the same as finding the coefficient of $x^4$ in the expansion of $$(1 + x + x^2 + x^3 + x^4)^7 (1 + x + x^2 + x^3 + x^4 + \cdots)^4 .$$ Since no term of degree $\geq 5$ can contribute, this is the same as the coefficient of $x^4$ in
$$(1 + x + x^2 + x^3 + x^4)^7 (1 + x + x^2 + x^3 + x^4)^4 = (1 + x + x^2 + x^3 + x^4)^{11} .$$
A: If you are still stuck ....
You are right to say that what you need is
$$
\text{ the coefficient of }\ x^{11}\ \text{in}\ 
x^7(1-x^5)^7\frac1{(1-x)^{11}}
$$
which is the same as
$$
\text{ the coefficient of }\ x^4\ \text{in}\ 
(1-x^5)^7\frac1{(1-x)^{11}}.
$$
Now the term $(1-x^5)^{7}$ can only contribute powers of $x^5$ which are too high; so we can ignore them and look for
$$
\text{ the coefficient of }\ x^4\ \text{in}\ 
\frac1{(1-x)^{11}}.
$$
By the Binomial Theorem this is just 
$$
\frac{11\cdot 12\cdot 13 \cdot 14}{1\cdot 2\cdot 3\cdot 4}.
$$
A: The following notation is sometimes convenient. We use $[x^n]$ to denote the coefficient of $x^n$ of a series.

In order to calculate the coefficient of $x^{11}$ we can write
\begin{align*}
[x^{11}]x^7(1-x^5)^7\frac{1}{(1-x)^{11}}
&=[x^4](1-x^5)^7\frac{1}{(1-x)^{11}}\tag{1}\\
&=[x^4]\sum_{j=0}^\infty\binom{-11}{j}(-x)^j\tag{2}\\
&=[x^4]\sum_{j=0}^\infty\binom{10+j}{j}x^j\tag{3}\\
&=\binom{14}{4}\tag{4}
\end{align*}

Comment:


*

*In (1) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

*In (2) we skip all powers of $x$ greater or equal $5$ in $(1-x^5)^7$ since they do not contribute to the coefficient of $x^4$. We also apply the binomial series expansion.

*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (4) we select the coefficient of $x^4$.
