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$$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$

I want to find the coefficients of $x^{10}$

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closed as off-topic by Zev Chonoles, O.L., Andres Caicedo, Asaf Karagila, Aang Jul 14 '13 at 10:34

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Please indicate your thoughts. (And the current tag is baffling.) –  Did Jul 13 '13 at 8:16
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What does "pls" mean? Last time I checked it was one of the two playlist formats in WinAmp. –  Asaf Karagila Jul 13 '13 at 8:37
    
'pls' means 'please'. –  Eckhard Jul 13 '13 at 9:40
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@Eckhard Probably sarcasm ;) –  dreamer Jul 13 '13 at 10:38

4 Answers 4

The not-quite-brute-force approach is feasible here. You can rewrite the expression as

$$\frac{1-x^7}{1-x}\cdot\frac{1-x^8}{1-x}\cdot\frac{1-x^9}{1-x}\;.$$

Now

$$\frac{1-x^7}{1-x}=1+x+x^2+x^3+x^4+x^5+x^6\;.$$

and in general

$$\frac{1-x^n}{1-x}=1+x+x^2+\ldots+x^{n-1}\;.$$

Thus, your expression is

$$\left(1+x+x^2+\dots+x^6\right)\left(1+x+x^2+\ldots+x^7\right)\left(1+x+x^2+\ldots+x^8\right)\;.$$

Before you combine like terms, each term in this product will have the form $x^i\cdot x^j\cdot x^k=x^{i+j+k}$, where $0\le i\le 6$, $0\le j\le 7$, and $0\le k\le 8$. Thus, you’re looking for the number of solutions to $i+j+k=10$, where $0\le i\le 6$, $0\le j\le 7$, and $0\le k\le 8$.

Finding the number of solutions in non-negative integers without any upper bounds is a standard stars-and-bars problem; accounting for the upper bounds requires a simple inclusion-exclusion argument. There are many examples of such calculation on the site; here is an answer that illustrates such a calculation.

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Corrected answer. –  Promise Jul 13 '13 at 10:06

$$(1-x^7)(1-x^8)(1-x^9)=1-x^7-x^8-x^9+\text{ powers of }x > 10$$

So, the coefficient of $x^{10}$ in $$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$

will be the coefficient of $x^{10}$ in $(1-x)^{-3}$

$- $ the coefficient of $x^{10-7}=x^3$ in $(1-x)^{-3}$

$- $ the coefficient of $x^{10-8}=x^2$ in $(1-x)^{-3}$

$- $ the coefficient of $x^{10-9}=x$ in $(1-x)^{-3}$

Using Generalized Binomial Theorem and Negative Binomial Series,

$$(1-x)^{-3}=1+3\cdot x+\frac{3(3+1)}{2!}x^2+\frac{3(3+1)(3+2)}{3!}x^3+\cdots +\frac{3(3+1)(3+2)\cdots(3+8)(3+9)}{10!}x^{10} +\cdots$$ if $|x|<1$

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@JonasMeyer, agreed and ammended –  lab bhattacharjee Jul 14 '13 at 9:46

Expand $(1-x)^{-3}=1+3x+6x^2+10x^3+...$. The coefficient of $x^{10}$ should now be easy to pick off.

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the idea of the question is

($1^n - b^n)= (1-b)(1 + b + b^2 + b^3...........)$

on solving, the $(1-x)$ terms in the numerator and denominator gets cancelled

and we get,

$(1 + x^2 + x^3 +....x^7)(1 + x^2 + x^3+....x^8)(1 + x^2 + x^3 +....x^9)$

now group the degrees to get $10$ $(x^10)$ :-

on finding the possible number of groups we get 60 different groups that can be formed and since the coefficient of each group is $1$,

the answer should be $60$

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Your highest exponents are off in your rewriting of the expression. The same method you suggest was correctly done in Brian M. Scott's earlier answer. –  Jonas Meyer Jul 13 '13 at 23:10

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