How many ways can $4$ people of each $n$ possible nationalities stand in a row with each person next to a fellow national? I am trying to solve the following problem:



The answer is this:



My doubt is the following - What are they counting in this problem? It's not clear to me by looking at the question and the answer. At first, I thought that we were counting this:

*

*Given $4$ places, in how many ways can place $4n$ people such that each person stands next to a fellow national?

After a bit of time thinking, it seems that they were counting:

*

*How many permutations of $4n$ people are there in which each person stands next to a fellow national?

But I tried to compute both quantities and my answer doesn't match. I want to know If I understood the problem wrong or if I am counting wrong.
 A: Your second interpretation is the correct one (in the sense that it matches the answer).
Maybe this helps.

Think of a row of $4n$ open spots that are to be filled up with persons.
So something like: $|-|-|-|-|-|-|\cdots\cdots\cdots|-|-|-|-|$
The condition that a person will always have a person of the same nationality next to him makes clear on forehand that we better interpret this as a row of $2n$ pairs of consecutive open spots.
So something like: $|--|--|--|\cdots\cdots\cdots|--|--|$
Every nationality supplies two pairs for filling up these pairs of consecutive open spots. Let's say they do that one by one.
For the first nationality supplying there are $\binom{2n}2$ choices possible. For the next nationality then $\binom{2n-2}2$ possibilities, et cetera.
That gives a total of $\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^n}$ possibilities.
For each nationality there are $4!$ ways to order the $4$ supplied persons.
So finally we arrive at $$\frac{(2n)!}{2^n}(4!)^n$$ possibilities.
