What does it mean to get rid of $x^{12}$ in the expression $ x^{12} (1-x^4)^6(1-x)^{-6} $ in the following context?

Question

What does the auctor mean when he says that he gets rid of $x^{12}$ in the following context ?

(...) For example,consider distributing $23$ toys among $6$ children such that no child gets more than $5$ or less than $2$ toys. Each child contributes a generating function of $ x^2 +x^3 +x^4 +x^5 $ , so the overall generating function is $$ ( x^2 +x^3 +x^4+x^5)^6 =x^{12} (1+x+x^2+x^3)^6 $$.

Using geometric series identities we have that the sum $ 1+x+x^2+x^3$ is equal to $ (1-x^4)/(1-x) $ , so our generating function is $ x^{12} (1-x^4)^6(1-x)^{-6} $. We get rid of the $ x^{12} $ by noting that the coefficient of $x^{23}$ in $ x^{12}(1-x^4)^6(1-x)^{-6}$ is the same as the coefficient of $x^{11}$ in $ (1-x^4)^6(1-x)^{-6} $ .

Then the author writes up the Binomial form of the expression.

Answer 1

The author is being a litte disingenuous. He doesn't really mean "get rid of the $x^{12}$", he means "we can disregard the factor $x^{12}$ in what we do next". What's of interest here is the value of the co-efficient of $x^{23}$ and he notes that that's the same as the co-efficient of $x^{11}$ (because the $x^{12}$ just increases the size of the powers of $x$ by 12, loosely speaking, without changing the co-efficients). So he can ignore the factor $x^{12}$ and write the rest as a binomial, and then at the end (for completeness) could bring the $x^{12}$ back in. This way solves him some writing.

Answer 2

It's not clear what you are asking, but note that your expression is $x^{12} \left( \dfrac {1-x^4} {1-x} \right)^6$ and since $1-x^4 = (1-x) (1+x+x^2+x^3)$, your expression will be $x^{12} (1+x+x^2+x^3)^6$, so you factor $x^{12}$ out of it by just... factoring it out!

Answer 3

Imagine that the coefficient of $x^{11}$ in the expression $\frac{(1-x^4)^6}{(1-x)^6}$ is $c$. That is there is a term $cx^{11}$. Now once you multiple the expression by $x^{12}$ then you are multiplying every term by $x^{12}$ including the term we just mentioned. So you get $cx^{11}\times x^{12}$ which gives $cx^{23}$ and we see that $c$ is the coefficient of $x^{23}$.

Answer 4

I don't know what you mean by getting rid of $x^{12}$ but you can simplify the expression by getting rid of $(1-x)^6$:

$$ \frac{x^{12} (1 - x^4)^6}{(1-x)^6} = \frac{x^{12} ((1 - x)(1+x)(1 + x^2))^6}{(1-x)^6} = x^{12} (1+x)^6 (1+x^2)^6.$$

Answer 5

I am not getting you but yes we can remove $x^{12}$ your expression can be written as $[(x^2).\frac{1-x^4}{1-x}]^6$ so $\frac{1-x^4}{1-x}=\frac{(1-x^2)(1+x^2)}{1-x}=(1+x)(1+x^2)$ so expression becomes $[(x^2+x^3).(1+x^2)]^6=[(x^2(1+x+x^2+x^3)]^6 $ so now you can do binomial expansion so we have reduced this expression and removed $x^{12} $