2008 AIME II Problem 12

On a pole with $b$ blue and $g$ green flags, the greens must occupy distinct gaps among the $b + 1$ gaps determined by the blues: $\binom{b+1}{g}$ ways

Sum $\binom{b+1}{g}\binom{11-b}{9-g}$ over $g$ using Vandermonde’s identity, then subtract the arrangements that leave a pole empty

Suppose the first pole gets $b$ blue and $g$ green flags, the second the remaining $10 - b$ blue and $9 - g$ green. On a pole with $b$ blue flags, the green flags must occupy distinct gaps among the $b + 1$ gaps around the blues, in $\binom{b+1}{g}$ ways. Temporarily ignoring the requirement that each pole be nonempty, the total is $$\begin{aligned} &\sum_{b=0}^{10} \sum_{g=0}^{9} \binom{b+1}{g} \\ &\quad {}\cdot \binom{11-b}{9-g} \\ &= \sum_{b=0}^{10} \binom{12}{9} \\ &= 11 \cdot 220 = 2420, \end{aligned}$$ where the inner sum collapses by Vandermonde’s identity, since $(b + 1) + (11 - b) = 12.$

The arrangements that leave a pole empty put all $19$ flags on one pole, in $\binom{11}{9} = 55$ ways for each choice of pole. Hence $$N = 2420 - 2 \cdot 55 = 2310,$$ and the remainder when $N$ is divided by $1000$ is $310.$