so I've come across a Probability generating function (pgf) question that requires me to find $f(21)$.The PGF is defined as follows.
$$G(t)=\left(\frac{t(1-t^6)}{6(1-t)}\right)^6$$
As mentioned before the task is to find $f(21)$.
I've used differenced of squares and difference of cubes to simplify $G(t)$ as follows:
$$G(t)= \frac1{6^6}[t + t^2 + t^3 + t^4 + t^5 + t^6]^6$$
Expanding the terms in the brackets $6$ times is not an efficient method! I'm struggling to find a way to get to the term $f(21)$ so that I can find the coefficient and hence the probability.
Any help is highly appreciated!
Best Answer
EDIT In the original post I omitted a sign (shown in $\color{red}{\text{red}}$ below.) I've corrected that and carried out the arithmetic.
I'll use the notation $[t^n]G(t)$ to mean the coefficient of $t^n$ in in $G(t)$ so that I understand you to be looking for $[t^{21}]G(t)$. We have $$ \begin{align} [t^{21}]G(t)&=[t^{21}]\left(\frac{t(1-t^6)}{6(1-t)}\right)^6\\ &=6^{-6}[t^{15}]\left(\frac{(1-t^6)}{1-t}\right)^6\\ &=6^{-6}[t^{15}]\sum_{n=0}^6\color{red}{(-1)^n}\binom{6}{n}t^{6n}\sum_{n=0}^\infty(-1)^n\binom{-6}{n}t^n\\ &=6^{-6}\sum_{n=0}^2\binom{6}{n}\binom{-6}{15-6n}(-1)^{15-5n}\\ &=6^{-6}\left(-\binom60\binom{-6}{15}+\binom61\binom{-6}{9}-\binom62\binom{-6}{3}\right)\\ &=6^{-6}\left(\binom60\binom{20}{15}-\binom61\binom{14}{9}+\binom62\binom{8}{3}\right)\\ &=\frac{4332}{46656}=\frac{361}{3888}\approx0.09285 \end{align}$$