Why doesn't combinatorics work here? A while ago I asked one-to-one in combinatorics and then using one-to-one
I'll repeat my answer here:

There are two distinguishable flagpoles, and there are $19$ flags, of which $10$ are identical blue flags, and $9$ are identical green flags. Let $N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when $N$ is divided by $1000$.

Then I tried to work it out here:

$$G B GBGBGB | BGBGBGBGBGB$$
I removed one blue flag between each green:
$$G G G G | BGGGGGB$$
There are:  $\binom{11}{9} = 55$ to arrange the blue/green. Then multiply by $10$ because of the divider of the poles.
$$= 55(10) = 550$$

But the answer is way off, what have I done wrong?
 A: It's a nice idea, but there are two problems. First, as Greg Martin says, sometimes different arrangements reduce to the same string. Secondly, some valid arrangements don't reduce to strings of the right type at all: GBBBBGBGBGBG|GBGBGBG is a valid arrangement, but if you reduce it you get three Bs, not two.
We can fix it, but it's more complicated. First, suppose that each pole has at least one green. In that case we can do the same conversion -- remove one B from between each pair of Gs -- except we only remove the B if the two Gs are on the same pole. So your example reduces to GGGGB|BGGGGGB. How many strings like this are there?

 This will always leave a string of 9 Gs, 3 Bs and 1 |, which can't be before or after all the Gs. So there are 8 different arrangements of just the Gs and |. For each of these there are $\binom{13}{3}=286$ ways to add the Bs. So that is 2288 in total. 

Finally, we need to deal with the arrangements where there is no G on one of the poles.

 If there are two Bs on that pole, there is only one option for the other pole. If there is only one, there are 10 places we can put the spare B on the other pole. So in total there are 22 such arrangements (since it can be either pole). Thus the overall total is 2310.

