I got the question on a midterm and got it wrong. I'd like to know where I went wrong. We were supposed to find the coefficient of $x^{15}$ of$$(1-x^2)^{-10}(1-2x^9)^{-1}$$
My answer
The only way to get $x^{15}$ is to select three $-x^2$'s and one $-2x^9$
There are $-10 \choose 3$ ways to choose three $-x^2$ with $${-10 \choose 3} = {10 + 3 - 1 \choose 3} = {12 \choose 3}$$ For each of the $12 \choose 3$ ways to get three $-x^2$'s, there's only one way to get a $-2x^9$.
The -1 from the $-x^2$ and the -2 from the $-2x^9$ are part of the coefficients, and so the coefficient of $x^{15}$ is $$2{12 \choose 3}$$
Is this the correct answer? If not, where did I go wrong? Thanks
You're correct the only way to make $x^{15}$ out of $-x^2$ and $-2x^9$ is to use three of the former and one of the latter. Therefore, using the binomial series, the coefficient should be
$$\binom{-10}{3}(-x^2)^3\,\cdot\,\binom{-1}{1}(-2x^9)^1. $$
Now,
$$\binom{-10}{3}=\frac{-10(-10-1)(-10-2)}{3\cdot2\cdot1}=-220, \qquad \binom{-1}{1}=-1 $$
(since $\binom{n}{1}=n$ identically), so the term is $440x^{15}$.
In your calculations, you seem to have missed the negative sign in $\binom{-10}{3}=-220$, and also didn't even reference $\binom{-1}{1}$ at all, which makes it seem like you've got two sign errors that cancelled each other out to give you the correct answer.