How many ways to arrange the flags? 
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 $1000$.

This is a tricky problem to be honest. 
Let $|$ distinguish the two flagpoles.
I tried arranging it as:
$$G B GBGBGB | BGBGBGBGBGB$$
$$G G G GB | BGGGGGB$$
There are:  $\binom{12}{3} = 220$ to arrange the blue/green. Then multiply by $11$ because of the divider of the poles.
$$= 220(11) = 2420$$
And this multiplication by $11$ takes care of the at least one flag on pole condition.
Then why is this the wrong answer?
 A: Here's a simple way to tackle the problem.
Blues on a flagpole can range from 0 to 10.
When 0 blue is on a flagpole (which means 1 green is there), there will be ${11\choose 8}$ ways to place the greens, else there will be ${12\choose 9}$ ways to place the greens.
Thus # of arrangements = $2\cdot{11\choose 8} + 9\cdot{12\choose 9}= 2310$
and remainder on dividing by 1000 = 310 
A: We don't change the number of arrangements if we stipulate that there are instead 12 blue flags and 9 green flags, and each flag must be topped with a blue flag. The reason is that we can go back and forth by adding/removing a blue flag from the the top of each flag pole.
Now the advantage of this change is that now every green flag is guaranteed to have a blue flag after it. So instead of 12 blue and 9 green, we have 3 blue and 9 [Green-Blue] chunks. So we can arrange these 12 symbols in any order, and then make a divide at any of the 11 gaps. This can be done in: $\binom{12}{3}\cdot 11$ ways. But we might have created an empty flagpole by isolating a blue flag on one end (and then removing it). This can be done in $\binom{11}{2}$ ways (on each side). So:
$$\binom{12}{3} \cdot 11 - 2 \cdot \binom{11}{2} = 2310$$
A: Another method would be to put one flagpole on top of the other.
Then we have 10 blue flags and 9 green flags arranged in order, and there are two possibilities:
1) If the flags where the flagpoles meet are not both green, we have an arrangement where no 2 green flags are consecutive; so there are $\binom{11}{9}$ ways to choose the 9 gaps for the green flags (from the 11 gaps created by the blue flags), and then
18 choices for separating the flags to form the two flagpoles.
2) If the flags where the flagpoles meet are both green, then there are 11 ways to select the gap for these two flags and $\binom{10}{7}$ ways to choose the gaps for the 7 remaining green flags (from the 10 remaining gaps). 
Therefore there are $\displaystyle 18\binom{11}{9}+11\binom{10}{7}=11(90+210)=2310$ such arrangements,
and the remainder when dividing by 1000 is 310.
