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I'm trying to find the coefficient of $x^{20}$ in $$(x^{2}+⋯+x^{6} )^{5}$$

My steps are

$$=x^{10}(1+⋯+x^{4} )^{5}$$

$$=\left(\dfrac {1-x^5} {1-x}\right)^{5} x^{10}$$

$$= (1-x^5)^5 * (1-x)^{-5} x^{10}$$

After that I am stuck, can someone show me how to use generating functions to finish this monster of a question? Please do not suggest using the multinomial theorem, I am learning generating functions right now so it would completely defeat the purpose of my learning if I used it :(. I'm thinking this is a convolution, but I don't know how it would look like multiplying two sequence together..Please, help is very much needed!

The answer is ${14\choose 10} - {5\choose 1} \times {9\choose 5} + {5\choose 2} $

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  • $\begingroup$ The equalities clearly don't hold. Are those typos? $\endgroup$ – user26486 Mar 15 '15 at 21:46
  • $\begingroup$ Yeah, just edited again, and made it mroe simple. $\endgroup$ – Belphegor Mar 15 '15 at 21:48
  • $\begingroup$ 14c2 - is it actually ${14 \choose 2}$? $\endgroup$ – HEKTO Mar 15 '15 at 22:12
  • $\begingroup$ I'm sorry, made a huge typo, it's actually 14c10 $\endgroup$ – Belphegor Mar 15 '15 at 22:13
  • $\begingroup$ what is the letter "c" meaning? $\endgroup$ – HEKTO Mar 15 '15 at 22:15
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what you are after if the coefficient of $x^{10}$ in the expansion $$\begin{align}[x^{10}](1-x)^{-5}(1-x^5)^5 &=[x^{10}] \left( 1 + {5 \choose 1}x + {6 \choose 2 }x^2 + \cdots \right)\left(1 - {5 \choose 1}x^5 + {5 \choose 2}x^{10} + \cdots\right)\\&={5 \choose 2} - {9 \choose 5} {5 \choose 1}+ {14 \choose 10}. \end{align} $$

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  • $\begingroup$ What happen to finding x^20 though? Does x^10 that was dragged out just become the new coefficient to find ? $\endgroup$ – Belphegor Mar 15 '15 at 22:34
  • $\begingroup$ @user3718584, you already pulled the $x^{10}$ out. $\endgroup$ – abel Mar 15 '15 at 22:39

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