Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna.  In how many ways can we choose 6 starters if at most one of the triplets is in the starting lineup?
We can add together the number of lineups with one triplet and with no triplets.   The number of lineups with no triplets is $\binom{11}{6} = 462$, since we must choose 6 starters from the 11 remaining players.  When one triplet is in the lineup, there are $3\cdot \binom{11}{5} = 1386$ options.  So the total number of lineups with at most one triplet is $1386 + 462 = \boxed{1848}$.