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Question

I cannot make sense of the expansion hint in ii) and how that could relate to solving this problem, and I would like an explanation of how these 2 hints are useful, preferably further hint before a solution.

(I solved the first part of the question through combinations & permutations and got an answer of 369)

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The answer to part (ii) is the same as the answer to "find the coefficient of $x^6$ in the expansion of $$(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7+9x^8-\cdots)^3$$" where inside the bracket is the infinite series $$1-2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7+9x^8-\cdots =\sum_{n=0}^\infty(-1)^n(n+1)x^n=\frac1{(1+x)^2}.$$

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  • $\begingroup$ I see that you can then do $$(\frac1{(1+x)^2})^3$$ to obtain the final answer, but then how would you compute $$\frac1{(1+x)^6}?$$ $\endgroup$ – LHC2012 Jun 1 '19 at 4:22
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    $\begingroup$ @LHC2012 Binomial theorem? $\endgroup$ – Angina Seng Jun 1 '19 at 4:23
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    $\begingroup$ It seems I never done binomial theorem with negative powers, so I may have to revise on that. Thank you. $\endgroup$ – LHC2012 Jun 1 '19 at 4:24

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