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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?

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1 Answer 1

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Your answer is right, your book is wrong. Here are two alternative ways to prove it:

  1. 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 $\binom{11+3-1}{3-1}$.
  2. Expand and look at the $x^{11}$ term.
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    $\begingroup$ Wow! Interesting. For future readers, this is #5 from Section 6.2 of Applied Combinatorics by Alan Tucker. $\endgroup$ Sep 24, 2017 at 3:43
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    $\begingroup$ @user3724404 The mistake doesn't seem to be listed in the errata. Perhaps you should email the author to let him know! $\endgroup$ Sep 24, 2017 at 4:25

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