$n$ goblins and $2n$ elves should be set into series We have $2n$ elves and n goblins. Elves are beautiful and everyone different than others. Goblins
They are ugly and all the same. We set the whole company in a $3n$-element series.
Goblins feel uncomfortable in the environment of elves, and therefore every goblin must have at least one another goblin as a neighbor. In how many ways can this company be set?
My approach
I want to count situations when goblins don't pass the condition.  So all possibilities for goblins are equal to $\binom{3n}{n}$ 
There are two more edge cases: 

*

*$1$ goblin is on the edge of series: $ \binom{2}{1}\binom{3n-2}{n-1}$

*$1$ goblin is on the edge of series: $ \binom{3n-2}{1}\binom{3n-3}{n-1}$
So goblins can be set on $$ \binom{3n}{n} - \binom{2}{1}\binom{3n-2}{n-1} - \binom{3n-2}{1}\binom{3n-3}{n-1} $$ ways.
Elves can be set on just $(2n)!$ ways. So the result should be:
$$ (2n)! \cdot \left( \binom{3n}{n} - \binom{2}{1}\binom{3n-2}{n-1} - \binom{3n-2}{1}\binom{3n-3}{n-1} \right) $$
but it doesn't work. If we take $n=2$ my pattern fails...
 A: Edit: I was not clear on saulspatz's answer, so I modified it a bit. Looking back, it appears the only difference was saulspatz was assuming there were $2n$ goblins and $n$ elves while I assumed the opposite, which I now realize was due to a typo in the title of this post. Anyway, here is my solution (very similar to saulspatz):
You have 2n elves. Between the elves are 2n+1 "gaps" where goblins can go. Any gap can have 0 or 2+ goblins put into it. First, we choose the gaps that have 2+ goblins in them. We can have anywhere from 1 through $\left\lfloor \dfrac{n}{2}\right\rfloor$ nonempty gaps.
This is the $g$ that saulspatz was discussing. There are $$\dbinom{2n+1}{g}$$ ways to choose which gaps are nonempty.
In each nonempty gap, we place 2 goblins. Since we started with $n$ goblins, there are now $n-2g$ goblins left to distribute to the $g$ bins we have available. By the Stars and Bars method, there are:
$$\dbinom{g+n-2g-1}{g-1} = \dbinom{n-g-1}{g-1}$$
ways to distribute the goblins to the gaps.
So, the final answer would be:
$$(2n)!\sum_{g=1}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \dbinom{2n+1}{g}\dbinom{n-g-1}{g-1}$$
If $n=2$, then the two goblins must be together, so there are $5!$ ways to order them. My formula gives: $$4!\dbinom{5}{1}\dbinom{0}{0} = 5!$$
If $n=3$, if there are 3 goblins, they must all be together because any smaller group will force a single goblin to be alone. Therefore, there are 3 goblins, plus 6 elves, which gives $7!$ ways to order them. This formula gives:
$$6!\dbinom{7}{1}\dbinom{1}{0} = 7!$$
For $n=4$, either all goblins are together ($9!$ ways to order them) or there are two groups of two goblins ($8!\dbinom{9}{2}$ ways to order them). This gives a total of $$5\cdot 9!$$
Plugging into my formula, you have:
$$8!\left(\dbinom{9}{1}\dbinom{2}{0} + \dbinom{9}{2}\dbinom{1}{1}\right) = 5\cdot 9!$$
This is looking really good!
A: HINITS
If there are $g$ groups of goblins, $1\leq g\leq n$, we need to figure out how many ways we can distribute $2n$ balls in $g$ bins with at least $2$ balls in each bin.  So, put $2$ balls in each bin, and then distribute the remaining $2n-2g$ balls.  By stars and bars there are $${2n-g-1\choose g-1}$$ ways to do this.
Now we have $g+1$ places to put the $n$ elves, and the middle places must each have at least one elf.  Agains, starts and bars gives the answer.  Since the order of the elves matters, we must multiply by $n!$.
Can you take it from here?    
