simple combinatorics - where is my mistake In the olympic games we want to organize 8 flags on 8 poles, 4 US flags, 2 french flags, 2 german flags.
We want to know how many combinations are there where a US flag is adjacent to a french flag.
Obviously the number of overall combinations is $\frac{8!}{4!2!2!} = 420$.
Now let's glue together an american flag and a french flag together. that gives us overall 7 flags. 3 america, 1 french, 2 german, 1 our new "flag block".
that's $\frac{7!}{3!1!2!1!} = 420$, but we need to double it since we have the internal combination of our block ("US France" or "France US"), so overall we have $2 \times 420 = 840$ combinations. But that makes no sense since overall we have $420$ combinations. where is the mistake
 A: You’re overlooking the fact that most of the acceptable arrangements have more than one instance of a U.S. flag adjacent to a French flag, and you’re counting those arrangements more than once each. It’s easier to count the arrangements in which a U.S. and a French flag aren’t adjacent and then subtract from the total. The point is that each French flag must have either the end of the row, the other French flag, or a German flag on each side of it.


*

*We can put the two French flags together and surround them with the German flags.  

*We can put the two French flags together at one end of the line with a German flag as ‘insulation’.  

*We can put a French flag at one end, followed by a German flag, the other French flag, and the other German flag.


Anything else will result in a U.S. flag next to a French flag. It’s not too hard to work out the number of arrangements in each of these three cases.
A: In addition to Brian M. Scott's answer, here's a clear example to show your multiple counting:
Consider the arrangement:
USA France USA Germany USA Germany France USA
How many times does your model count this?
1: The USA France block is the first two flags.
2: The France USA block is the second and third flag.
3: The France USA block is the last two flags.
So you count this one configuration three times. 
In particular, it multiply counts each arrangement once for each France-USA adjacency it contains.
