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Find the number of 5-combinations of the multi-set {4*a, 4*b, 4*c} using generating functions.

Because I am finding combinations I know that I have to use ordinary generating functions and not exponential generating functions.

I can first represent this as $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4)$$ $$=(1+x+x^2+x^3+x^4)^3$$

I'm not sure where to go from here though- I would greatly appreciate any advice or help! Thanks.

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  • $\begingroup$ You could expand that polynomial (though that's probably not the cleanest way to do it) $\endgroup$ – Dennis Meng Dec 5 '13 at 18:32
  • $\begingroup$ After expanding it, what would I do to find the exact number of 5-combinations? $\endgroup$ – user2553807 Dec 5 '13 at 19:42
  • $\begingroup$ As already mentioned in the answer, you'd be looking for the coefficient to $x^5$. (Remember, the exponents in generating functions mean something; it's not just a magic formula) $\endgroup$ – Dennis Meng Dec 5 '13 at 19:45
  • $\begingroup$ If I'm looking at the coefficient of x^5 wouldn't it be -3? Can you explain your comment a little more or help me find more information on that. Thanks!! $\endgroup$ – user2553807 Dec 5 '13 at 21:44
  • $\begingroup$ Given that the polynomial has zero negative signs, I haven't a clue where you got a negative sign from. $\endgroup$ – Dennis Meng Dec 5 '13 at 21:48
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This is the coefficient of $x^5$ in the expansion of $(1+x+x^2+x^3+x^4)^3 = (1-x^5)^3/(1-x)^3 = (1-3x^5 +3x^{10} -x^{15})\displaystyle\frac{1}{(1-x)^3} $. Can you find a taylor series expansion for the second term in the product?

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