jmc

It's tempting to do this problem using casework, but there's an easier way. There are a total of $\left(\genfrac{}{}{0ex}{}{14}{6}\right)=3003$ ways to select a lineup with no restrictions. Of those 3003 lineups, the only ones that don't satisfy the given condition are ones that contain all three triplets. There are $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$ of these, since once we insert the three triplets in the lineup we have 3 spots left to fill using the remaining 11 players. Subtracting gives us our answer: $3003-165=\boxed{2838}$ possible starting line ups.