Our school's girls volleyball team has 14 players, including a set of 3 triplets: Missy, Lauren, and Liz.  In how many ways can we choose 6 starters if the only restriction is that not all 3 triplets can be in the starting lineup?
It's tempting to do this problem using casework, but there's an easier way. There are a total of $\binom{14}{6}=3003$ 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 $\binom{11}{3}=165$ 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.
2838
Level 4
Counting & Probability
Problem: Our school's girls volleyball team has 14 players, including a set of 3 triplets: Missy, Lauren, and Liz.  In how many ways can we choose 6 starters if the only restriction is that not all 3 triplets can be in the starting lineup?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we first determine the total number of ways to choose 6 starters from 14 players without any restrictions. This can be calculated using the combination formula C(n, k), where n is the total number of players and k is the number of players to choose. Thus, we calculate C(14, 6).

Next, we need to account for the restriction that not all 3 triplets can be in the starting lineup. To do this, we will first calculate the number of ways to choose 6 starters that include all 3 triplets. If all 3 triplets are included, we need to choose 3 more players from the remaining 11 players (14 total players minus the 3 triplets). This is calculated as C(11, 3).

Now, we can find the total number of valid combinations by subtracting the restricted combinations (where all triplets are included) from the total combinations. Therefore, the number of valid combinations is given by:

Total valid combinations = C(14, 6) - C(11, 3).

Next, we compute the values of these combinations. 

C(14, 6) = 14! / (6! * (14 - 6)!) = 14! / (6! * 8!) = 3003.

C(11, 3) = 11! / (3! * (11 - 3)!) = 11! / (3! * 8!) = 165.

Now, we substitute these values into our equation for total valid combinations:

Total valid combinations = 3003 - 165 = 2838.

Thus, the final answer is 2838.<|END_OF_LATENT|> The final answer is $2838$.