<s>[INST] Find the coefﬁcient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7+...)^3.$ This is #5 from Section 6.2 of Applied Combinatorics by Alan Tucker:

Find the coefﬁcient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7+...)^3.$

The books answer: $C(11+3−1,11)−C(3,1)×C(5+3−1,5)$.
I'm not sure where the second term comes from? Here is my work:
$$\begin{array}{l}
(x^2+x^3+x^4+x^5+x^6+x^7+\ldots)^3\\
=x^6(1 + x + x^2 + \ldots)^3\\
=x^6\dfrac{1}{(1-x)^3}\\
=x^6\left(1 + C(1+3-1,1)x + C(2 +3-1,2)x^2+C(3+3-1,3)x^3+ \ldots \right)
\end{array}$$
So, the coefficient of $x^{17}$ is really going to be the coefficient of $x^{11}$. So our answer is: $$C(11+3-1,11).$$
But that's only half the book answer stated above. Why?
 [/INST] Your answer is right, your book is wrong.  Here are two alternative ways to prove it:


*

*The coefficient of $x^{11}$ is the number of ways to put 11 $x$'s in 3 baskets.  This is the "stars and bars" problem, and the answer is $\binom{11+3-1}{3-1}$.

*Expand and look at the $x^{11}$ term.
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