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?
Best Answer
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