Using the probability generating function, find the probability of the sum of 5 throws of a die being 15.
I got the answer as $\frac{1}{6^5} \left( \frac{t(t^6-1)}{t-1}\right)^5 $. When I expand this on Wolfram Alpha, I got the coefficient of $t^{15}$ as $651$. So is the answer $\frac{651}{6^5} = \frac{651}{7776} = \frac{217}{2592}$ . I am confused because some people told me I can't use $t^5$ in the upper equation to expand.
Yes, the generating function that gives the probability to obtain a certain sum of $5$ throws of a die is $$f(t)=\left(\frac{t+t^2+t^3+t^4+t^5+t^6}{6}\right)^5=\left(\frac{t(1-t^6)}{6(1-t)}\right)^5.$$ I confirm that by extracting the coefficient of $t^{15}$ we get $217/2592$: $$[t^{15}]f(t)=\frac{1}{6^5}[t^{10}]\left(\frac{1-t^6}{1-t}\right)^5=\frac{1}{6^5}\sum_{k=0}^1(-1)^k\binom{5}{k}\binom{-5}{10-6k}\\= \frac{1}{6^5}\left(\binom{-5}{10}-5\binom{-5}{4}\right) =\frac{651}{6^5}=\frac{217}{2592}.$$