The Fiberglass Menagerie

The number of ways that n objects can be split into k groups of $\frac{n}{k}$ objects each is

$\frac{n!}{\left(\left(\frac{n}{k}\right)!\right)^k k!}.$

So the question boils down to which of the following is greatest:

$\frac{20!}{\left(10!\right)^2 2!}, \quad \frac{20!}{\left(5!\right)^4 4!}, \quad \frac{20!}{\left(4!\right)^5 5!}, \quad \frac{20!}{\left(2!\right)^{10} 10!}.$

Equivalently, we just need to determine which of

$\left(10!\right)^2 2!, \quad \left(5!\right)^4 4!, \quad \left(4!\right)^5 5!, \quad \left(2!\right)^{10} 10!$

is the least. So to do this, we'll compare them in pairs, then compare the lesser of each pair. Since

$\frac{(4!)^5 5!}{(5!)^4 4!}=\frac{4!4!4!4!4!5!}{5!5!5!5!4!}=\frac{4!}{5^3}=\frac{24}{125},$

this tells us that $(4!)^5 5!<(5!)^4 4!$ . And since

$\frac{(2!)^{10} 10!}{(10!)^2 2!}=\frac{2^9}{10!}=\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2},$

it's clear that $(2!)^{10} 10!<(10!)^2 2!$ . Now for the finals!

$\frac{(4!)^5 5!}{(2!)^{10} 10!}=\frac{4^5 3^5 2^5 1^5 5!}{2^5 2^5 10!}=\frac{2^5 3^5 5!}{10!}=\frac{6 \cdot 6 \cdot 6 \cdot 6 \cdot 6}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6},$

so $(4!)^5 5!$ is the least, which means that $\frac{20!}{(4!)^5 5!}$ is the greatest. So the answer is $\boxed{\text{Five cages, 4 animals per cage}}$ .