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?
If all triplets are in the starting lineup, we are choosing the 3 remaining starters from 11 players, which can be done in $\binom{11}{3} = \boxed{165}$ ways.
To solve this problem, we can break it down into cases and use the principle of inclusion-exclusion.
First, let's consider the total number of ways to choose 6 starters from the 14 players without any restrictions. This can be calculated using combinations:
C(14, 6) = 14! / (6! * (14 - 6)!) = 3003
Next, let's calculate the number of ways to choose 6 starters where all 3 triplets are included. We need to choose 3 players from the remaining 11 players (excluding the triplets):
C(11, 3) = 11! / (3! * (11 - 3)!) = 165
Since we want to exclude this case, we subtract it from the total:
3003 - 165 = 2838
However, we need to consider the case where exactly 2 of the triplets are included in the starting lineup. There are 3 ways to select the 2 triplets and 9 remaining players to choose 4 more players:
3 * C(9, 4) = 3 * (9! / (4! * (9 - 4)!)) = 630
Next, we need to consider the case where all 3 triplets are excluded from the starting lineup. In this case, we need to choose 6 players from the remaining 11 players:
C(11, 6) = 11! / (6! * (11 - 6)!) = 462
Now, let's subtract these cases (where both triplets are included and where all triplets are excluded) from the total:
2838 - 630 - 462 = 1746
Therefore, there are 1746 ways to choose 6 starters if the only restriction is that not all 3 triplets can be in the starting lineup.
According to AoPS, the answer is 2838
990 according to AoPs Alcumus
2838
answer is 165
There are 11 players who can be chosen at will
If no triplet is chosen, you have 11C6 ways to do it
If 1 is chosen, you have 11C5*3C1 ways to do it
If 2 are chosen, you have 11C4*3C2 ways to do it
so add them up