Counting the number of possible arrangements of animals in how many ways can we arrange 5 elephants, 4 zebras and 3 monkeys in a row such that no species is arranged together.
It's a bit vague, I know so here are some examples for "good" and "bad" arrangments
eeeeezmzmzmz - "bad" because all the elepants are together
ememzzzzemee - "bad" because all the zebras are together
ezmezmezmeze - "good"
 A: We assume that the only forbidden configurations have all elephants together, or all zebras, or all monkeys. We will assume that animals of the same species are indistinguishable. If they are distinguishable, just multiply the answer we get by $5!4!3!$. 
To count, we use Inclusion/Exclusion. There are $\binom{12}{5}$ ways to choose the elephant spots, and for each way there are $\binom{7}{4}$ ways to choose the zebra spots. Multiply and simplify. We get $\frac{12!}{5!4!3!}$. We can also express this as a multinomial coefficient. 
We must subtract from $\frac{12!}{5!4!3!}$ the number of bad arrangements.
Let's count how many arrangements have the elephants all together. To do this,  put the elephants in a bag, viewing the elephants as a single object. We have $8$ objects, which can be arranged in $\frac{8!}{1!4!3!}$ ways. Similarly, there are $\frac{9!}{5!1!3!}$ ways in which the zebras are together, and $\frac{10!}{5!4!1!}$ ways in which the mokeys are.
But if we add these, we will overcount the situations in which for example the elephants and the zebras are together. If we think of these as single objects, there are $5$ objects, and $\frac{5!}{1!1!3!}$ ways to arrange them. You can write down similar expressions for the ways elephants are together and monkeys are together, also for the ways zebras and monkeys are.  
But if we add these and subtract from  $\frac{8!}{1!4!3!}+\frac{9!}{5!1!3!}+\frac{10!}{5!4!1!}$, we will undercount the bads, since we will subtract once too many times the $3!$ arrangements in which all species are together.
Now you have all the ingredients to find the answer. 
