<s>[INST] What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do. 
I would really appreciate any help
 [/INST] You can pick up an $x^5$ from one term and a $1$ from every other term 17 different ways.
You can pick up an $x^4$ from one term, an $x$ from a second term in $\dfrac{17!}{15!}$ ways.
You can get $x^3$ and $x^2$ in $\dfrac{17!}{15!}$ ways.
You can get $x^3$, $x$, $x$ in $\dfrac{17!}{1!2!14!}$ ways.
You can get $x^2, x^2, x$ in $\dfrac{17!}{1!2!14!}$ ways.
You can get $x^2,x,x,x$ in $\dfrac{17!}{1!3!13!}$ ways.
And finally, you can get $x,x,x,x,x$ in $\dbinom{17}{5}$ ways.
Show that this is exhaustive, and it should give you the correct answer.
Final tally:
$$17+2\cdot \dfrac{17!}{15!} + 2\cdot \dfrac{17!}{2!14!} + \dfrac{17!}{3!13!} + \dfrac{17!}{5!12!} = 20,349$$
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