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I would like to know a way to solve the problem. How can we divide n identical marbles into k distinct with each pile having at most w marbles. I have seen a solution due to Brian Scott on the problem using inclusion-exclusion but I would like one via generating functions. In particular how can i solve with $n=30,k=10$ and $w=4$??

Asinomás
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  • The particular case seems trivial, you need all $10$ piles to have the full complement of $3$ marbles. In general you could look for the coefficient of $x^n$ in the expansion of $(1+x+x^2+\cdots+x^w)^k$. – Macavity Dec 28 '13 at 18:03
  • I mean w=4 and n=30. My bad. – Asinomás Dec 28 '13 at 18:06
  • Well, then you have the answer. Its the coefficient of $x^{30}$ in $(1+x+x^2+x^3+x^4)^{10} = \dfrac{(1-x^5)^{10}}{(1-x)^{10}}$, the second expression may be a little easier to evaluate the coefficient by hand (you would need only binomial theorem to find and add some $7$ terms)... – Macavity Dec 28 '13 at 18:19
  • Could you please ellaborate why they are the same things in an answer. I would be willing to accept an explanation – Asinomás Dec 28 '13 at 18:53
  • OK will do, though I guess you could figure out easily. – Macavity Dec 28 '13 at 19:07

1 Answers1

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The choice in each pile can be represented as $(1+x+x^2+x^3+x^4)$, corresponding to no marbles, 1 marble etc. all the way to 4 marbles. As there are $10$ piles, we have the possibilities equal to the coefficient of $x^{30}$ in $(1+x+x^2+x^3+x^4)^{10}$ - corresponding to each arrangement of the piles, there will be a contribution to the coefficient of $x^{30}$ from some one term in each of the factors.

Now we can expand that out and find the result, or note that $1-x^5 = (1-x)(1+x+x^2+x^3+x^4)$, so we are searching for the same coefficient in the series expansion of $\dfrac{(1-x^5)^{10}}{(1-x)^{10}}$, which may be simpler to compute as the product of two sums below:

$$(1-x^5)^{10} = \sum_{k=0}^{10} \binom{10}{k}(-1)^k x^{5k} = 1-10 x^5+45 x^{10}-120 x^{15}+210 x^{20}-252 x^{25}+210 x^{30}-120 x^{35}+45 x^{40}-10 x^{45}+x^{50} $$

Of these, only the first $7$ terms can contribute to the coefficient of $x^{30}$.

$$(1-x)^{-10} = \sum_{k=0}^{\infty} \binom{-10}{k}(-1)^k x^{k} = 1+\frac{10}{1!}x+\frac{10\cdot 11}{2!}x^2+...\frac{10\cdot 11\cdot 12\cdot13\cdot14}{5!}x^5+... $$

Note that we need only the terms of degree $5k$ in the second series. Cross multiplying the terms which contribute to $x^{30}$ you have your answer.

Macavity
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