Arrange 15 animals in 15 cages Five tigers, five lions and five cheetahs are going to be hosted in $15$ consecutive cages in the zoo. Due to some restrictions in the zoo regulations, we cannot put the tigers in the $5$ leftmost cages, the lions in the $5$ middle cages and the cheetahs in the $5$ rightmost cages. Can you calculate all possible arrangements?
If all animals were the same, there would be $15!$ different ways. Now that we have $5$ of every kind, if there were no restrictions from the zoo regulations, we would have $\frac{15!}{5!5!5!}$. I don't know though how to apply the restrictions :(
 A: Let's turn it into letters, animals $a,b,c$ and cages $A,B,C$, where $a$ can't be in $A$. 
We have to decide on the number $n$ of animals $b$ to place in $A$. This can be anything from $0$ to $5$. So we later will have to sum over all these options.
As soon as you place $n$ animals $b$ in cage $A$, $(5-n)$ animals $b$ need to be placed in $C$. The rest of $C$ will be filled with $n$ times $a$, so the number $n$ applies to all three sets of cages!
Within each set of cages, we have $5\choose {n}$ possible placements, so the toal will be:
$$\sum_{n=0}^{5} 
{5\choose n}^3
$$

Sidenote to address one of your comments:
If you just look at the first $N$ cages with indistinguishable animals, $$\sum_{n=0}^{N} 
{N\choose n} = 2^N$$ what you observed in one of your comments.
A: I'm assuming you cannot place any of the tigers in any of the leftmost cages and so on. For the moment the $5$ members of each animal family are undistinguishable. I'll follow Pieter's notation with animals $a, b$ and $c$ and cages $A, B$ and $C$ with forbidden $a$ in $A$, $b$ in $B$ and $c$ in $C$.
Start placing the five $a$'s. You have $10$ places available so you have ${10 \choose 5} = 252$ arrangements. These can be cathegorized in three cases:
$I)$ all $5$ $a$ animals in the same cage ($B$ or $C$);
$II)$ $4$ animals in one cage and $1$ in the other;
$III)$ $3$ in one and $2$ in the other.
Each of these cases counts $\times 2$, since the roles of $B$ and $C$ can be swapped. Let's see how many arrangements correspond to each case:
$I)$ there is only one way to place $5$ undistinguishable $a$ animals in, say, $B$. Considering the $\times 2$ factor (you could place them in $C$) we have the first $2$ of said $252$;
$II)$ $4$ $a$ animals can be placed in $5$ ways in $B$ (easy if you think about the $5$ possibilities for the "hole") and the remaining one can be placed in $5$ places in $C$, so there are $25$ arrangements. Invert $B$ and $C$ and we have other $50$ of the $252$;
$III)$ $3$ $a$ animals can be placed in $5 \choose 3$ ways in $B$ and the remaining $2$ in $5 \choose 2$ ways in $C$. This gives ${5 \choose 3}\cdot {5 \choose 2}=\frac{5!}{3!2!}\cdot \frac{5!}{2!3!}=10\cdot 10=100$ possibilities. And we have the final $200$ arrangements of the $252$.
Now we have to place another species. Let's consider first the one that has been more constrained. Looking at each case we have:
$I)$ consider again that all $a$ animals were placed in $B$, then all $c$ animals must be placed in $A$. Finally all animals $b$ must go in $C$. Only one possible arrangement for each of the $2$, so this case $(I)$ brings exactly $2$ arrangements;
$II)$ in $B$ there is one "hole", and since it cannot be filled with a $b$ we must place there a $c$. Now we have $5$ ways to place the remaining $4$ $c$'s in $A$. Finally we are forced to place the $b$'s in the only possible way. So, $5$ alternatives for each of the $50$ make case $(II)$ give $250$ valid arrangements;
$III)$ again $2$ holes in $B$ must be filled with $c$ animals. The remaining $3$ can be placed in $A$ in ${5 \choose 3}=10$ ways. Again $b$'s take the places that are left. So $10\cdot 200$ contribute with $2000$ arrangements within case $(III)$.
The final total amount of arrangements that comply with the given restriction is:
$$2 + 250 + 2000 = \fbox{2252}$$
If you want the animals to be distinguishable, you have to consider the permutations within each species, so multiply the result by $5!\cdot5!\cdot 5!$, and we're done!
