What is the coefficient of $x^{27}$ for $f(x) = \Big(\dfrac{1+x^{10}}{(1-x)^5}\Big)^{2}$ ?
So i can work the math a little bit to get -
$f(x) = \Big(\dfrac{1+2x^{10} + x^{20}}{(1-x)^{10}}\Big)$ $\iff f(x) = (1+2x^{10} + x^{20}) \cdot (\sum_\limits{n=0}^{\infty}\binom{10+n-1}{n}x^n)$
So i want to sum $1$ for the left with $x^{27}$ for the right, $2x^{10}$ for the left with $x^{17}$ from the right and $x^{20}$ from the left with $x^7$ from the right.
$ = 1 \cdot \binom{36}{27} + 2\cdot\binom{26}{17} + 1\cdot\binom{16}{7}$
Is it done right ?