Coefficient of $x^{27}$ in $(1-x^{10})^6/(1-x)^6$. It's been a while since I have been playing with these so excuse me if it is too obvious.
How can I represent 
$$ f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6} $$
as sum of powers of $x$
I am especially interested in the coefficient in front of $x^{27} $ in that sum. 
The book I am reading gives this coefficient as obviously being 
$${32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$
but I don't know where this comes from. 
Many thanks in advance. 

First from the geometric series
$$
\frac{1}{1-x}=\sum_{n=0}^\infty x^n
$$
combined with repeated differentiation yields that
$$
\frac{1}{(1-x)^6}=\sum_{n=0}^\infty\binom{n+5}{5} x^n\tag{1}
$$
(for $|x|<1$, if you don't want to deal with formal series). The binomial theorem implies that
$$
(1-x^{10})^6=1-6x^{10}+\binom{6}{2}x^{20}-\binom{6}{3}x^{30}+\binom{6}{4}x^{40}-\binom{6}{5}x^{50}+x^{60}\tag{2}.
$$
We want the coefficient of $x^{27}$ in the product of (1) and (2). We get an $x^{27}$ term  from the products $1\times x^{27}$, $x^{10}\times x^{17}$ and $x^{20}\times x^{7}$. In particular the coefficient is then
$$
\binom{27+5}{5}-6\binom{17+5}{5}+\binom{6}{2}\binom{7+5}{5}.
$$
