Imagine there are $3$ distinct items: $A,B,C$.
(i) The number of ways of selecting one or more items from $3$ distinct items is:
$${3\choose 1}+{3\choose 2}+{3\choose 3}=2^3-1=7 \ \ \ (\text{note:} \sum_{k=0}^n {n\choose k}=2^n).$$
Indeed, the outcomes are:
$$A,B,C,AB,AC,BC,ABC.$$
Now imagine there are $3$ identical items: $A,A,A$.
(iv) The number of dividing $3$ identical items among $2$ persons, each of them who can receive 0,1,2, or more items is:
$${3+2-1\choose 2-1}={4\choose 1}=4 \ \ \ (\text{explained below})$$
Indeed, the outcomes are:
$$\{AAA,0\}, \{AA,A\}, \{A,AA\}, \{0,AAA\}.$$
Explanation: let $x_1,x_2$ be the number of identical items the two persons receive, respectively. Then, the equation is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
1-method: Stars and bars. Consider the number of stars as number of items to be given to a person and a bar to switch from one person to another person. For example, several examples:
$$\begin{align}**|* &\ \ (\text{the first person gets $2$, the second gets $1$})\\
|*** &\ \ (\text{the first person gets $0$, the second gets $3$})\\
*|** &\ \ (\text{the first person gets $1$, the second gets $2$})\\
***| &\ \ (\text{the first person gets $3$, the second gets $0$})
\end{align}$$
Note that it is basically a combination of $4$ symbols (items) taken $1$ (bar) or $3$ (stars) at a time:
$${4\choose 1}={4\choose 3}=4.$$
2-method: Generating functions. The equation to be solved is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
Let the equation
$$(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3)=x^{3}$$
represent the number of items (indicated on the exponents) the two persons (indicated by the brackets) can receive. For example:
$$x^0\cdot x^3=x^3 \ \ (\text{the first person gets $0$, the second gets $3$})\\
x^2\cdot x^1=x^3 \ \ (\text{the first person gets $2$, the second gets $1$})\\
x^3\cdot x^0=x^3 \ \ (\text{the first person gets $3$, the second gets $0$})\\
x^1\cdot x^2=x^3 \ \ (\text{the first person gets $1$, the second gets $2$})\\$$
Now if we consider the sum of $x^3$ we get $4x^3$. So, the problem reduces to finding the coefficient of $x^3$ in the expansion of the left-hand side:
$$\begin{align}[x^3](1+x+x^2+x^3)^2&=[x^3]\left(\frac{1-x^4}{1-x}\right)^2= \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)\\
&=[x^3](1-x^4)^2(1-x)^{-2}= \ \quad \quad \quad \quad \quad \quad (2)\\
&=[x^3]\sum_{k=0}^2 {2\choose k}(-x^4)^k\cdot \sum_{k=0}^{\infty} {-2\choose k}(-x)^k= \ \ (3)\\
&=[x^3]{2\choose 0}(-x^4)^0\cdot {-2\choose 3}(-x)^3= \quad \quad \quad \quad (4)\\
&={2+3-1\choose 3}= \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (5)\\
&={4\choose 3}.\end{align}$$
where:
(1) inside brackets: the sum of four terms of GeomProg.
(2) simple algebra.
(3) negative binomial series.
(4) considering only the terms, whose exponents add up to $3$.
(5) negative binomial theorem.
Best Answer
The power series expression of $\frac{1}{1-x}$ is $\sum_{n = 1}^\infty x^n$, so you can rewrite what you have as $$ (x^{48} - 6x^{40} + 15x^{32} - 20x^{24} + 15x^{16} - 6x^8 + 1)(1 + x + x^2 + x^3 + \cdots)^6. $$
Now, you want to see what ways you can get a term of $x^{25}$. Let $p(n,k)$ be the number of ways to partition the integer $n$ into at most $k$ parts. A term of $x^{25}$ can arise from $-20x^{24} \cdot p(1,6)x$, $15x^{16} \cdot p(9,6)x^9$, $-6x^8 \cdot p(17,6)x^{17}$, or $1 \cdot p(25,6)x^{25}$.
The partition numbers are coming into play because we need to know how many ways we can form a particular power of $x$ by choosing a term from each of the 6 copies of $\frac{1}{1-x}$, but that's precisely the same as asking how many ways you can partition that particular integer exponent into at most 6 parts.