Find the coefficient of $x^{19}$ in the generating function development of $f(x) = \dfrac{(1-x^{5})^5}{(1-x)^{10}}$
My Attempt:
Let $\mathbf{(I)} \ (1-x^5){^5}$, then $(1-x^5){^5} = \sum_{k=0}^{5} \binom{5}{k} \left((-1)^{k}\cdot x^{5k}\right)$
And let $\mathbf{(II)} \ \left(\frac{1}{1-x}\right)^{10} = \left(\sum_{l=0}^{\infty}{\left(x^{l}\right)^{}}\right)^{10} = \sum_{l=0}^{\infty} \binom{l+9}{9}\cdot{x^{l}}$
I'd like to find all values of $k,l$ for which we'll get $x^{19}$.
as $k=0,l=19:x^{5\cdot(0)}\cdot x^{19}=x^{19}$, and the coefficient of $x^{19}$ is $\binom{28}{9}$
as $k=1,l=14:x^{5\cdot(1)}\cdot x^{14}=x^{19}$, and the coefficient of $x^{19}$ is $-5\cdot \binom{23}{9}$
for $k=2,l=9:x^{5\cdot(2)}\cdot x^{9}=x^{19}$, and the coefficient of $x^{19}$ is $10\cdot\binom{18}{9}$
for $k=3,l=4:x^{5\cdot(3)}\cdot x^{4}=x^{19}$, and the coefficient of $x^{19}$ is $-15\cdot \binom{13}{9}$
Finally, the coefficient of $x^{19} = \binom{28}{9} -5\binom{23}{9} +10\binom{18}{9} -15\binom{13}{9}$
Is that True?
You are almost correct! Since $\binom{5}{3}=10$ (not $15$), it should be $$[x^{19}]f(x) = \binom{28}{9} -5\binom{23}{9} +10\binom{18}{9} -\color{red}{10}\binom{13}{9}=3300000.$$