Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters.  There are 12 players on the team.  Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup.  How many different starting lineups are possible?  (The order of the players in a basketball lineup doesn't matter.)
Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup.  The order in which the players are chosen doesn't matter, so the answer is $$ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \boxed{120}. $$
120
Level 3
Counting & Probability
Problem: Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters.  There are 12 players on the team.  Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup.  How many different starting lineups are possible?  (The order of the players in a basketball lineup doesn't matter.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of different starting lineups, we first recognize that Ace and Zeppo are already included in the lineup, leaving us with 3 additional spots to fill. Since there are 12 players total and 2 are already chosen, we have 10 players remaining to choose from. 

The task now is to select 3 players from these 10. The order of selection does not matter, so we will use combinations to calculate the number of ways to choose 3 players from 10. The formula for combinations is given by:

C(n, k) = n! / (k! * (n - k)!)

In this case, n is 10 (the remaining players) and k is 3 (the players we need to select). 

Calculating this gives us:

C(10, 3) = 10! / (3! * (10 - 3)!) 
          = 10! / (3! * 7!) 
          = (10 * 9 * 8) / (3 * 2 * 1) 
          = 720 / 6 
          = 120

Thus, the total number of different starting lineups possible, including Ace and Zeppo, is 120.<|END_OF_LATENT|> The final answer is $120$.