<s>[INST] 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. 
 [/INST] As an alternative recall that
$$(1-x^{10} )=(1-x)(1+x+x^2+\ldots+x^9)$$
therefore
$$f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6}=(1+x+x^2+\ldots+x^9)^6$$
then we can evaluate the coefficient for $x^{27}$ by the stars and bars method by the equivalent problem, discussed here, of the numbers of way to distribute $b=27$ balls in $c=6$ container with no more than $n=9$ balls in any container that is
$$N(b,c,n)=\sum_i(-1)^i\binom{c}i\binom{b+c-1-i(n+1)}{c-1}=$$
$$=\binom{27+6-1}{5}-\binom{6}{1}\binom{27+6-1-(9+1)}{5}+\binom{6}{2}\binom{27+6-1-2(9+1)}{5}=$$
$$={32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$
</s>