Find the mean, variance and $f(21)$ of this if it is a p.g.f. $$G(t) = t^{11}e^{9t-9}$$
Find the mean, variance and $f(21)$
Please help me with find $f(21)$ as easily as possible.
For the mean and variance:
$$G'(t) = 11t^{10}e^{9t-9}+9t^{11}e^{9t-9}$$
$$G^{(2)}(t) = 11(10) t^9e^{9t-9} + 11(9)t^{10}e^{9t-9} + 9(11) t^{10}e^{9t-9} + 9(9)t^{11}e^{9t-9}$$
$$\mu = G'(1) = 11+9 = 20$$
$$\mu_{(2)} = G^{(2)}(1) = 110+99 +99 + 81 = 389$$
$$\mu_2' = \mu_{(2)} + \mu = 389 + 20 = 409$$
$$\sigma^2= \mu_2 = \mu_2'-\mu^2 = 409 - 20^2 = 9$$
Now for $f(21)$
$$f(21) = \frac{G^{(21)}(0)}{21!}$$
Okay now surely there must be a better way of finding the $21^{st}$ derivative other than deriving $21$ times but I don't know how I would go about finding the $r^{th}$ factorial moment for this one.

$$G(t) = \frac{t^{11} e^{9t}}{e^9} = \sum_{n=0}^\infty \frac{9^nt^{n+11}}{n!e^9}$$
Now $$G^{(21)}(t) = \sum_{n = 10}^\infty\frac{9^n(n+11)\cdots (n-9)t^{n-10}}{n!e^9}$$
Hence in $0$ the only term that survives is with $n=10$, so 
$$G^{(21)}(0) = \frac{9^{10}\cdot 21\cdot 20 \cdots 1}{10!e^9} = \frac{9^{10}21!}{10!e^9}$$
$$f(21) = \frac{G^{(21)}(0)}{21!} = \frac{9^{10}}{10!e^9} \sim 0.118$$
