Final answer:
There are 84 different starting lineups possible for the Las Vegas Gamblers basketball team.
Step-by-step explanation:
To find the number of different starting lineups, we first need to determine the number of choices for each position. There are 4 guards to choose from for the first guard position. After selecting the first guard, there are 3 guards left to choose from for the second guard position. Similarly, there are 7 forwards to choose from for the first forward position, and 6 forwards left to choose from for the second forward position. Finally, there are 4 centers to choose from for the center position.
We can calculate the total number of different starting lineups by multiplying these choices together: 4 choices for the first guard position * 3 choices for the second guard position * 7 choices for the first forward position * 6 choices for the second forward position * 4 choices for the center position = 4 * 3 * 7 * 6 * 4 = 1008.
However, we have to account for the fact that the order of the players in the lineup does not matter. Therefore, we need to divide the total number of lineups by the number of ways to arrange the players within the lineup. The lineup consists of 2 guards, 2 forwards, and 1 center, so there are 2! (2 factorial) ways to arrange the guards, 2! ways to arrange the forwards, and 1! way to arrange the center. 2! = 2 * 1 = 2, and 1! = 1.
Therefore, the final number of different starting lineups is 1008 / (2 * 2 * 1) = 84