Why do we subtract the combination?

[Amad27]

Problem:
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 $1000$.

My solution attempt was:
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Let $|$ distinguish the two flagpoles.

I tried arranging it as:
Quote:
$$G B GBGBGB | BGBGBGBGBGB$$

Using the one-to-one correspondence idea, take out the Blues in-between.

$$G G G GB | BGGGGGB$$

There are: $\binom{12}{3} = 220$ to arrange the blue/green. Then multiply by $11$ because of the divider of the poles.

$$= 220(11) = 2420$$

So each pole has at least one flag (since we only multiplied by 11).

But the real solution also subtracts:

$$2420 - 2\binom{11}{2}$$

But I don't understand why.

 

[Valkarie]

Why are we subtracting $2\binom{11}{2}$?A:The $2\binom{11}{2}$ is the number of arrangements with two green flags on one pole and none on the other. We are subtracting this because these arrangements are not allowed in the problem.