I was looking at an example in my textbook where it was asked to find in how many ways can a police captain distribute 24 rifle shells to four police officers so that each officer gets at least three shells, but not more than eight. Using generating functions: ${f(x)=(x^3+x^4+x^5+...+x^8)^4}$, and we will be looking to find the coefficient of ${x^{24}}$ in $f(x)$
We know that $f(x)= x^{12}((1-x^6)/(1-x))^4$
Therefore the answer will be the coefficient of $x^{12}$ in the expansion of $(1-x^6)^4*(1-x)^{-4}$
Which in turn is equal to $A$:
$[1-\dbinom{4}{1}x^6+\dbinom{4}{2}x^{12}-\dbinom{4}{3}x^{18}+x^{24}]*[\dbinom{-4}{0}+\dbinom{-4}{1}(-x)+\dbinom{-4}{2}(-x)^2+...]$
Up to this point I perfectly understand the problem and the proposed solution; but the textbook goes on saying that the above expression is equivalent to $B$: $[\dbinom{-4}{12}(-1)^{12}-\dbinom{4}{1}\dbinom{-4}{6}(-1)^6+\dbinom{4}{2}\dbinom{-4}{0}]$
I do not understand how they went from $A$ to $B$
I know that $C$: $[\dbinom{-4}{12}(-1)^{12}]$ is the coefficient of $x^{24}$ in: $x^{12}[\dbinom{-4}{0}+\dbinom{-4}{1}(-x)+\dbinom{-4}{2}(-x)^2+...]$
But I do not seem to make the link between $C$ and the following subtraction and addition, namely:
$[C-\dbinom{4}{1}\dbinom{-4}{6}(-1)^6+\dbinom{4}{2}\dbinom{-4}{0}]$
If $(1-x^6)^4$ has $x^k (12\ge k\ge 0)$ term then we select coefficient of $x^{12-k}$ in $(1-x)^{-4}$ Multiplying both coefficients results in the coefficient of $x^{k} + x^{12-k} = x^{12}$ which is the required coefficient we need. Other terms will have contribution $0$ in finding the coefficient of $x^{12}$.